Note: When studying the following material, make sure to completely redraw the figures on your notebook and write the formulas as you proceed. This will help you learn better. Make sure to always draw horizontal fraction bars.
The current, I through a wire is defined as the change in charge, Δq per change in time, Δt at any cross-sectional area of that wire. We may generally write:
The SI unit for I is of course Coul./sec called Ampere or Amp or simply A .
Example 1: A 5.00-Amp current flows through a wire. Calculate the amount of electric charges that cross any section of that wire in 20.0 minutes.
Solution: Since I = Δq / Δt , we may cross-multiply to solve for Δq. We get: Δq = I Δt or ,
Δq = (5.00Amps)(20.0*60sec.) = 6000 Coul., better to write as (6.00x103 Coul.)
Besides Coulombs, another unit for the amount of electricity (Δq) is Ampere-hour (Ah). It is easy to see that:
1 Ah = 3600 Coul. This reason is : 1 Ah = 1 (Coul./s)(3600 s) = 3600 Coul.
Example 2: A car battery has been under charge for 40.0 hours at an average current of 0.80 Amps. How much electricity in (Ah) is stored in it? How much is that amount in (Coul.)?
Solution: I = Δq /Δt ; Δq = IΔt ; Δq = (0.80 A)( 40.0 h) = 32.0 Ah
Δq = (32.0)( 3600) Coul. = 115,000 Coul.
For any linear electric device, the voltage, V across the device divided by the current, I through the device is a constant. That constant is called the resistance, R of the device. Mathematically,
where in SI, the unit of R is volts/amp called "Ohms" with the symbol Ω , pronounced "Omega."
Example 3: Calculate the resistance R of a light bulb that allows a current of 2.5 Amps to flow through it when connected to a 12-volt battery.
Solution: The light bulb resists toward the passage of the current through it. We use a zigzag line to show the resistance of an electric device. We also use two parallel lines one shorter and one longer as a symbol for a battery. The longer line means the positive pole. The following figure is self-explanatory:
Solution: ( Write with horiz. fraction bars.)
R = V / I = (12 V) / ( 2.5 A)
R = 4.8 Ohms, or
R = 4.8 Ω
When key K is turned on, the circuit is closed and the current, I flows making the light bulb go on.
Test Yourself 1:
1) The electric current, I, is defined as (a) the # of gallons of water crossing a section of a pipe per second (b) the # of Coulombs of electric charge crossing a section of a wire per second or per unit of time. (c) neither a nor b.
2) The unit for electric current is (a) ampere. (b) Coulombs per second. (c) both a & b. click here
3) If through a cross-section of a wire, 480 Coulombs of electric charge passes per minute, the currents is (a) 8.0 Amps. (b) 480 Amps. (c) 480 mA.
4) If through a cross-section of a wire, 7200 μCoul. of electric charge passes every hour, the current is (a) 2.0A. (b) 2.0μA. (c) mA. click here
5) If through a cross-section of a wire, 108 μCoul. of electric charge passes in 1/2 hour, the current is (a) 6.0mA. (b) 6.0μA. (c) 60nA.
6) 2Amps my be written as (a) 2,000mA. (b) 2000,000μA. (c) both a & b. click here
7) 350mA may be written as (a) 350,000A (b) 0.35A. (c) 3.5A
8) 3600Coul. is (a) 1.0 Ah (b) 10.Ah (c) 0.10Ah.
9) If a current of 5.0A passes through an electric device for 6.0hours, the amount of electric charge, Q flown through that device is (a) 30.Ah (b) 108,000Coul. (c) both a & b. click here
10) If a battery goes under charge for 24h at an average current of 1.5A, the amount of electric charge, Q accumulated in the battery is (a) 20.Ah. (b) 36Ah. (c) neither a nor b.
11) Ohm's Law states that (a) the voltage-to-current ratio of a resistor remains constant. (b) the voltage-to-current ratio of a resistor is not a constant. (c) the coulombs-to-current ratio is constant. click here
12) Ohm's law may be written as (a) V = RI. (b) I = V/R. (c) both a & b.
13) The voltage across a resistor is 10.0volts and the current through it 2.50A. Its resistance is (a) 0.250Ω. (b) 4.00Ω. (c) 25.0Ω.
14) The current through a resistor is 2.30A and the voltage across it 9.20V. Its resistance is (a) 4.00Ω. (b) 0.400Ω. (c) 21.2Ω. click here
15) The current through a 2.0Ω-resistor is 3.0A. The voltage across it is (a) 6.0V. (b) 0.67V. (c) 1.5V.
16) The voltage across a 4.5Ω-resistor is 27V. The current through it is (a) 120A. (b) 0.17A (c) 6.0A.
17) If V is known, and either R or I is to be calculated, then (a) to find R, V must be divided by I. (b) to find I, V must be divided by R. (c) both a & b. click here
18) If R and I are known, to find V, (a) R and I must be multiplied because V = RI. (b) R and I must be divided by each other somehow because V = RI. (c) both a & b.
19) In Fig. 1, if Vbattery = 12.0V, the voltage across the resistor, R (a) is 12.0V, because there is only one resistor in the circuit and all the battery voltage drop across that single resistor. (b) would be 12.0V anyway even if other resistors R1, R2, R3, ... were present in series with R. (c) neither a nor b. click here
20) Although the ammeter in Fig. 1 has some small resistance itself ; however, in calculations, it is neglected and therefore the voltage across R is (a) 12.0V. (b) slightly more than 12.0V. (c) neither a nor b.
Problem: A 12-V battery is connected to a 2.4-Ω car light bulb.
21) The current through the bulb is (a) 29A. (b) 5.0A. (c) 0.20A. click here
22) If the resistance in the circuit is doubled (by adding another 2.4-Ω resistor in series with the first one), the current will be (a) halved to 2.5A. (b) doubled to 10.0A. (c) does not change.
23) Based on the previous question, for a fixed voltage, if the resistance is doubled, the current (a) doubles. (b) remains unchanged. (c) halves. click here
24) Read the problem's statement again. If the 2.4-Ω resistor is replace by a 0.80-Ω one (the resistance cut to 1/3), the current (a) nine folds. (b) does not change. (c) triples.
Problem: Suppose there is a source that can supply a fixed current of 3.0A to a user regardless of the resistance of the user. Answer the following questions: click here
25) If this 3.0-A current is passing through an 8.0Ω-resistor, the voltage across it is (a) 2.67V. (b) 0.375V. (c) 24V.
26) If the 8.0Ω-resistance is doubled to 16Ω, the voltage across it (a) doubles to 48.0Ω. (b) halves to 12V. (c) does not change. click here
27) From the previous question, one may conclude that for a fixed current, if the resistance doubles, then the voltage (a) halves. (b) doubles. (c) triples.
Problem: Suppose there is a source that can supply a fixed voltage of 24V to a user regardless of the resistance of the user. Answer the following questions: click here
28) When the resistance connected to this source is 1.2Ω, the current is (a) 20.A. (b) 0.02A. (c) 26.4A.
29) If the resistance is changes to 12Ω ( increased by a factor of 10), the current (a) also increases by a factor of 10 to 200A. (b) decreases by a factor of 10 to 2.0A. (c) remains unchanged.
30) From the above, one may conclude that for a fixed voltage, if the resistance increases by a certain factor, the current then (a) decreases by the same factor. (b) increases by the same factor. (c) remains unchanged. click here
An Excellent Link to Be Used at Several Section of This Chapter:
Resistance of a Wire:
Running electric current through a wire is similar to running water in a pipeline. The longer the pipe, the smaller its cross-sectional area, and the rougher it is inside, the more difficult and forceful is the pushing of water through it. An electric wire acts in a similar manner. The longer the wire, L, the smaller its cross-sectional area, A, and the less conductive it is ( ρ = resistivity), the greater resistance, R it has toward an applied voltage in allowing a current to pass through it. The dependence of R on L, A, and ρ is given in the following formula:
where ρ, pronounced "rho", depends on the material. For copper ρcu = 1.70x10-8Ωm at 20ºC. Table 20.1 gives the value of for a few selected materials:
ρ ( Ωm )
|α ( ºC ) -1||Element||
ρ ( Ωm )
|α ( ºC ) -1|
|Aluminum||2.82 x 10-8||4.29 x 10-3||Carbon||3.5 x 10-5||-5.0 x 10-4|
|Copper||1.70 x 10-8||6.80 x 10-3||Germanium||4.5 x 10-1||-5.0 x 10-2|
|Iron||10 x 10-8||6.51 x 10-3||Silicon||2.2 x 10 2||-7.0 x 10-2|
|Nickel||7.8 x 10-8||6.0 x 10-3||
|Platinum||10 x 10-8||3.93 x 10-3||Glass||1012|
|Tungsten||5.6 x 10-8||4.5 x 10-3||Rubber||1015|
Example 4: Calculate the resistance of a power line (copper) 425km long and 1.27cm thick (1/2 inch.).
Solution: The diameter of the circular wire is D =1.27cm. Its radius is therefore r = 0.635cm = 6.35x10-3m.
R = ( ρL) /A ; R = (1.7x10-8 Ωm)( 425000m) / [π (6.35x10-3)2 m2] = 57.0 Ω
Example 5: Calculate the resistance of a thin copper wire 0.100mm thick and 50.0m long, This wire is only 50.0m ling but it is as thin as hair.
Solution: The diameter of the circular wire is D =0.100mm. R = 0.0500mm = 5.00x10-5m.
R = ( ρL) /A ; R = (1.7x10-8 Ωm)(50.0m) / [π (5.00E-5)2 m2] = 108 Ω
Power dissipation in a resistor :
Any electric device that pulls some current, I from a source and has some voltage, V across it, consumes some electric energy per unit of time. This means that each electric device has a power, P.
It is easy to show that (electric power) is equal to (Voltage) times (Amperage) . Simply, P = V I .
The SI unit for power is [P] = [ V ][ I ] = ( Joule / Coul )(Coul / sec ) = (Joule / sec).
Joule / sec is called Watt as was learned in Physics I. This means that watt = ( volt )( amp ).
Example 6: A light-bulb that works with 120 Volts pulls a current of 0.83 Amps from a wall electric outlet. Find its power.
Solution: P = V I = (120 V )( 0.83 A ) = 100 watts.
Example 7: A 60.0-watt light-bulb is connected to a wall outlet, V = 120 volts. Calculate (a) the current it pulls and (b) its resistance.
Solution: (a) P = VI ; therefore, I = P / V ; I = (60.0 watts) / (120 volts) = 0.50 Amps.
(b) V = RI ; R = V / I. ; R = (120 volts) / (0.5 Amps) = 240 Ω.
Another Version of P = V I
Since V = R I, We may write P = ( R I )( I ) or P = R I2.
Example 8: A current of 5.0Amps is flowing through an electric heater with a resistance of 40.0 Ω. Find (a) its electric power in watts, (b) its daily ( 24 hours ) energy consumption in ( Joules ), and (c) the daily heat energy in calories it gives to the room. Note: 1 cal = 4.186 J
Solution: (a) : P = RI2 = (40.0 Ω )( 5.0 A )2 = 1000 watts ( 3 sig. fig.)
(b) P = (Work / time) ; therefore, Energy = ( P )( time) , Here Energy = (1000. watts )(24)(3600s)
Energy = 86,400,000 Joules (c) Energy = (86,400,000 / 4.186) = 2.1x107 calories
A commercial Unit for Energy (kwh):
It is easy to show that kilo-watt-hour (kwh) is a unit of energy. Since P = W/t, work or energy (W) can be solved in terms of power (P) and time (t) by cross-multiplying the formula P = W/t. This cross-multiplication yields:
W = Pt ; expressing P in kilo-watts and t in hours, gives us energy or work, W in kwh.
Example 9: A water heater works at 240.V and pulls 12.5A. Find (a) its power in watts and kilowatts. If this heater is on 12h a day during a 30-day month, find (b) its consumption in kwh. If electric energy sells for 6.5 cents per kwh, calculate (c) the cost per month of this heater.
Solution: (a) P = VI ; P = ( 240.V )( 12.5A ) = 3000 watts ; P = 3.00 kw
(b) W = Pt ; W = ( 3.00kw)( 360h) = 1080 kwh
(c) Cost = ( 1080 kwh )( $ 0.065 / kwh ) = $ 70.2
Test Yourself 2:
1) The resistance, R of a wire depends (a) directly on its length, L. (b) directly on its resistivity, ρ. (c) directly on its cross-sectional area. (d) directly on the reciprocal of its cross-sectional area (1/A). (e) a, b, & d. click here
2) The resistivity, ρ of a resistor depend on (a) the material of the resistor. (b) the length of it. (c) the cross-sectional area of it. (d) a, b, and c.
3) Comparing two wires of the same thickness and the same material, (a) the longer wire has a greater resistance. (b) the longer wire has a smaller resistance. (c) both have the same resistance regardless of their length. click here
4) Comparing two wires of the same length and the same material, (a) the thicker wire has a greater resistance. (b) the thicker wire has a smaller resistance. (c) both have the same resistance regardless of their thickness.
5) Two wires A and B of the same length and the same material. The diameter of wire A is 1/3 of that of wire B. Comparing the resistances of wires A and B, we may write: (a) RA = (1/3) RB. (b) RA = 3 RB. (c) RA = 9 RB. click here
6) Two wires C and D of same length and same material, wire C has a diameter 1/6 that of wire D. Comparing the resistances of wires C and D, we may write: (a) RC = 36RD. (b) RC = 6RD. (c) RC = (1/6)RD.
7) The resistance of a copper power line 255.0 miles long (1mile = 1609m) and 2.00cm thick is (a) 45.3Ω (b) 22.2Ω. (c) 11.1Ω. click here
8) The resistance of a copper wire 41.0m (135 ft) long and 0.200mm thick is (a) 45.3Ω (b) 22.2Ω. (c) 11.1Ω.
9) The reason why the resistances of the two wires in the previous two questions are equal is that (a) the second wire is much shorter but also much thinner. (b) the first wire is much thicker but also much longer. (c) both a & b. click here
10) If a wire is passed through a narrow hole to where its diameter is reduced by a factor of 2, its length becomes (a) 2 times longer. (b) 4 times longer. (c) 6 times longer.
11) The reasoning for the previous question is that (a) when the diameter or radius is cut by half, the cross-sectional area is reduced by a factor of 4. (b) A = π r2. (c) the volume of the wire is kept constant in the process. If the cross-sectional area is reduced by a factor of 4, the length has to increase by a factor of 4 to keep the volume constant. (d) a, b, & c.
12) If a wire is passed through a narrow hole to where its diameter is reduced by a factor of 2, its resistance, R becomes (a) 16 times greater. (b) 4 times greater. (c) 8 times greater. click here
13) If a wire is passed through a narrow hole to where its cross-sectional area, A is reduced by a factor of 3, its resistance, R becomes (a) 3 times greater. (b) 9 times greater. (c) 6 times greater.
14) If a wire is passed through a narrow hole to where its length, L is increased by a factor of 3, its resistance, R becomes (a) 3 times greater. (b) 6 times greater. (c) 9 times greater. click here
15) The resistivity, ρ of conductors is of the order of (a) 10-8 Ωm. (b) 10-5 Ωm. (c) 10+12 Ωm.
16) The resistivity, ρ of semiconductor, germanium is of the order of (a) 10-8 Ωm. (b) 10 -1 Ωm. (c) 10+12 Ωm.
17) The resistivity, ρ of insulator, glass is of the order of (a) 10-8 Ωm. (b) 10 -1 Ωm. (c) 10+12 Ωm. click here
18) Power is defined as (a) the work done per unit of time. (b) the energy given off per unit of time. (c) the energy consumed per unit of time. (d) a, b, & c.
19) The electric power, P may be written as (a) P = VI. (b) P = RI*I. (c) P = RI2. (d) a, b, & c.
20) If the voltage across a resistor is 15 volts and the current through it is 2.0A, the power dissipation in it is (a) 7.5 watts. (b) 30. watts. (c) 60. watts. click here
21) The voltage across a dirt devil is 120 volts and the current through it is 4.0A, the power dissipation in it is (a) 480 watts. (b) 90. watts. (c) 560. watts.
22) The voltage across a dirt devil is 120 volts and the current through it is 4.0A, the electric resistance it shows while in use is (a) 480 ohms. (b) 160 ohms. (c) 30. ohms. click here
23) The power of a dirt devil with an in-use resistance of 30.Ω that allows a 4.0-A current to flow through it is (a) 30(4)2watts (b) 480 watts. (c) both a & b.
24) An alternate version of the power formula, P = VI, is (a) P = RI2. (b) P = RI. (c) both a & b. click here
25) The voltage across an electric iron is 120V and the in-use electric resistance of it is 15Ω. The electric current it pulls from the socket and the power it dissipates as heat are (a) 10A and 1000watts. (b) 8.0A and 960watts. (c) 15A and 1800watts.
26) A light bulb is rated as 75 watts designed for a 120V use. The current it pulls from the socket and its in-use resistance are: (a) 0.625A and 192Ω. (b) 1.25A and 384Ω. (c) 1.60A and 75Ω. click here
27) P = VI may be written as (a) P = V (q / t) (b) P = Vq / t (c) P = energy / t (d) P = energy per unit of time. (e) a, b, c, and d. click here
28) Since power = energy / time, solving for energy, we get: (a) energy = power X time. (b) Ue = P t (c) both a & b.
29) The power of a light bulb is 150watts. The electric energy, Ue, it consumes in 30.0s is (a) 5.0 J (b) 4500J. (c) 0.20J
30) An electric motor is rated as 2.50hp. Its power in watts is (a) 1890 watts. (b) 1380watts. (c) 790 watts.
31) If a 2.50-hp motor is on for 2.00 minutes, the energy, in Joules, it consumes is (a) 3960J. (b) 990J. (c) 227000J.
32) The power in kw (kilo-watt) of a 2300watt electric motor is (a) 2,300,000kw (b) 2.3kw (c) neither a nor b.
33) The power in kw of a 150-watt light-bulb is (a) 0.15kw (b) 150,000kw. (c) neither a nor b. click here
34) The power in kw of a 4.00-hp electric pump is (a) 4.00kw (b) 6.00kw (c) 2.98kw. click here
35) The electric energy, Ue, a 4.00-hp pump consumes in 2.0h, expressed in kwh, is (a)5.96kwh. (b)12.0kwh. (c)18kwh.
36) A 100.-w light bulb has been on for 25h. The consumed energy in kwh is (a) 2.5kwh. (b) 2500kwh. (c) 0.25kwh.
37) The energy a 40.0w bulb consumes in 30.0 days in Joules and kwh is (a)1200J, 288kwh. (b)1.04x108 J, 28.8kwh. (c) neither a nor b. click here
38) For a 6.5¢ / kwh enrgy price, the cost of leaving a 75.0-w light-bulb on for 10.0 days is (a) $1.77. (b) $1.50. (c) $1.17.
Now let us consider cases where there are more devices, resistors in a circuit. The resistors could be in series or in parallel.
Resistors in Series:
Devices ( resistors ) in a portion of a circuit are said to be in series if they experience the same current. The following diagram shows three resistors in series with a battery. We can show that if R1, R2, and R3 are in series, there is a single resistor Rt (called the total or the equivalent resistance) that can replace them all such that
Rt = R1 + R2 + R3
|In a series
circuit as shown:
1) The current, I , is the same in all resistors, but the voltage-drop across each resistor is different.
2) Positive current leaves the (+) pole of a battery. Going clockwise, along with the current shown, from the (-) pole to the (+) pole of the battery, the voltage (V) goes up. Then as the current flows through each of R1, R2, and R3, the voltage drops proportionally. The total voltage drop must be equal to the voltage jump produced by the battery. We can write:
Vb = V1 + V2 + V3. Also, V1= I R1 and V2= I R2 and V3 = I R3 ; Consequently ,
Vb = I R1 + I R2 + I R3. But Vb is used up for the total resistance Rt, and according to Ohm's law Vb = Rt I , or,
Rt I = I R1 + I R2 + I R3. simplifying :
Rt = R1 + R2 + R3
Example 10: In the above figure, let V = 20.0 volts, R1 = 25.0Ω, R2 = 50.0Ω, and R3 = 25.0Ω. Calculate Rt, I, V1, V2, and V3.
Solution: Rt = R1 + R2 + R3 = 100.0Ω. Using Ohm's Law: I = (V / R) = (20.0 volts / 100.0Ω ) = 0.200 Amps.
Now: V1 = I R1 = (0.200Amps)(25.0Ω) = 5.00 volts
V2 = I R2 = (0.200Amps)(50.0Ω) = 10.0 volts
V3 = I R3 = (0.200Amps)(25.0Ω) = 5.00 volts
Resistors in Parallel:
Resistors that are between the same two nodes experience the same voltage and form parallel connection. At one node the arriving current divides amongst the resistors and at the other node the currents join and form the main undivided current. The greater a resistor, the smaller the current its branch pulls. Parallel resistors divide the main current amongst themselves proportional to 1/R . The voltage for all resistors between two nodes is the same. The equivalent resistance, Rt or Req may be calculated from the formula:
In the figure shown, since there is no resistance between points d and a ( Connecting wires have negligible resistance) the voltages of points d and a are equal. So are the voltages of points b and c. This means that if we use a voltmeter to measure Vdc and Vab, we get the same readings. Vdc = Vab = Vbat . Since Vab is known, Vbat , we may calculate the currents in each branch by using the Ohm's law.
I1 = Vab / R1, similarly I2 = Vab / R2 and I3 = Vab / R3.
The voltage Vab = Vdc is the same for all resistors and is equal to the battery voltage (Vbat); therefore,
I 1 = Vbat / R1
I 2 = Vbat / R2
I 3 = Vbat / R3
Of course, the 3 resistors can be replaced by one resistor, Req or Rt such that I = Vbat / Rt
Since I = I 1 + I 2 + I 3 ; therefore,
(1 / R t ) = (1 / R1) + ( 1 / R2) + ( 1 / R3)
Example 11: In the following figure, find the equivalent resistance in each case.
Solution: In Fig. 1, the two resistors are in series and they simply add up to a greater resistance (15Ω ).
In Fig. 2, the two resistors are in parallel and their reciprocals add up to yield the reciprocal of the total that is smaller than the smaller resistor (6Ω ).
In Fig. 3, the two parallel resistors yield 8Ω. The 8Ω is then in series with the 9Ω . They add up to 17Ω.
In Fig. 4, each parallel section must be calculated first. The first set yields 18Ω, and the second set yields 8Ω. The 18Ω and the 8Ω are then in series and add up to 26Ω.
A Shortcut for Two Parallel Resistors:
There is a shortcut formula that calculates the equivalent of two parallel resistors quickly. The formula is:
Example 12: In the previous example, verify the parallel sets by using the above formula.
Solution: Should be done by students.
Example 13: In the above figure, let R1 = 30.0Ω, R2 = 6.00Ω, and R3 = 3.33Ω. Also let Vbat = 4.00 volts. Find Rt, I, I1, I2, and I3.
Solution: Rt = 1 / (1/R1 +1/R2 +1/R3 ) = 2.00Ω. Ohm's Law: I = Vbat / Rt ; I = 4 volts / 2Ω = 2.00Amp.
Vab = Vdc ;therefore, I 1 = Vab / R1 = 4 / 30 = 0.133A
I 2 = Vab / R2 = 4 / 6 = 0.667A
I 3 = Vab / R3 = 4 / 3.33 = 1.200A, Obviously : I1 + I2 + I3 = 2.00A
Mixed Series and Parallel Circuits:
There are many cases in which a parallel portion is in series with other portions. The other portions may also be comprised of series or parallel portions. A typical case is shown in the following example:
Example 14: In the figure shown, find the current through, voltage across, and power dissipation in each resistor:
Let R1 = 12 Ω, R2 = 60. Ω, R3 = 5.0 Ω, and V = 45 volts.
Rab = 1 / ( 1/ R1 + 1 / R2 ) = 10. Ω
Rac = Rab + Rbc = 10 + 5 = 15. Ω
I = Vac / Rac = (45 volts) / (15 ohms) = 3.0 amps
Vbc = ( I )( Rbc ) = 15 volts
Vab = Vac - Vbc = 45 - 15 = 30. volts
I 1 = Vab / R1 = 2.5 amps
I 2 = Vab / R2 = 0.50 amps
P1 = (V1)(I1) = 30x2.5 = 75 watts
P2 = 30x0.5 = 15 watts ; P3 = 15x3.0 = 45 watts
Example 15: In the figure shown, find the equivalent resistance between points a and b.
Solution: In the Problem, if a current flows from a to m, it has to divide at m, partially toward n, and partially downward. The part that arrives at n must also divide, partially through the 5Ω-resistor and partially through the 60Ω-resistor. The part that flows through the 5Ω-resistor must also flow through the 15 resistor as well, and therefore, the 5Ω- and 15Ω-resistors are in series. The first step is to simply add the 5Ω- and 15Ω-resistors to get an equivalent of 20Ω resistance as shown in Fig.1. In Fig. 1, Since n is a dividing point, the 20Ω and 60Ω resistors are in parallel. They can be replaced by their equivalent resistance of (20*60)/(20+60) = 15Ω as shown in Fig. 2.
In Fig. 2, the 25Ω- and 15Ω-resistors are in series because whatever current flows through the 25Ω must also flow through the 15Ω-resistor. They add up to 40Ω as shown in Fig. 3. In Fig. 3 , the 40Ω- and the 120Ω resistors are in parallel because the current that arrive at m from a must divide. They can be replaced by their equivalent resistance of (120*40)/(120+40) = 30Ω as shown in Fig. 4. In Fig. 4, the 50Ω- and the 30Ω-resistors are in series because the flow through the 50Ω-resistor must also flow through the 30Ω-resistor. They add up to a final value of 80Ω.
Test Yourself 3: click here
1) Resistors in series experience (a) the same current. (b) the same voltage. (c) the same power dissipation.
2) Resistors in parallel experience (a) the same current. (b) the same voltage. (c) the same power dissipation.
3) Resistors in mixed series and parallel experience (a) the same current. (b) the same voltage. (c) neither a nor b.
4) Total resistance for a group of resistors in series is (a) greater than each individual resistance. (b) equal to the sum of the individual resistances. (c) both a & b. click here
5) Total resistance for a number of resistors in parallel is (a) greater than the greatest resistance. (b) smaller than the smallest resistance. (c) equal to the sum of the individual resistances.
6) The Rtotal or Req for a 120.Ω-, an 80.Ω-, and a 48Ω-resistor connected in series is (a) 96Ω. (b) 248Ω. (c) 280Ω.
7) The Rtotal or Req for a 120.Ω-, an 80.Ω-, and a 48Ω-resistor connected in parallel is (a) 96Ω. (b) 248Ω. (c) 24Ω.
8) R1 = 120.Ω and R2 = 80.Ω are in parallel and the result in series with R3 = 48Ω. Draw a figure and place the parallel set between points A and B, and the third resistor, R3, between points B and C. The overall resistance between points A and C is (a) 96Ω. (b) 248Ω. (c) 24Ω. click here
9) Between points E and F, a 6.00Ω- and a 30.0Ω-resistor are in parallel, and between points F and G, a 12.0Ω- and a 60.0Ω-resistor are also in parallel. The set between E and F is in series with the set in between F and G. Draw a diagram for the problem. The overall resistance between E and G is (a) 3.33Ω. (b) 108Ω. (c) 15.0Ω.
10) In Example 15, redraw the figure. Double the even numbers and triple the odd number in the main problem. The new equivalent resistance between points a and b is (a) 173Ω. (b) 273Ω, (c) 178Ω. click here
11) In Example 10, let V = 60volts, R1 = 50Ω , R2 = 150Ω, and R3 = 100Ω. Redraw the figure. Without looking at the solution, recalculate new values. The values of Rt, I, V1, V2, and V3 all to 3 sig. fig. are (a) 600Ω, 0.4A, 20V, 60V, and 40V. (b) 300Ω, 0.2A, 10V, 30V, and 20V. (c) 200Ω, 0.1A, 25V, 65V, and 45V.
Problem: Let a 24-volt battery be connected to following three resistors in parallel : R1 = 8Ω , R2 = 12Ω, and R3 = 6Ω. Draw a diagram for the problem. Answer the following questions: click here
12) We may say that the voltage across (a) R1 is 24volts. (b) R2 is 24volts. (c) R3 is 24volts. (d) a, b, & c.
13) The currents through the individual resistors are respectively, (a) 3A, 2A, & 4A. (b) 3A, 3A, & 3A. (d) neither a nor b.
14) The current that leaves the battery is (a) 24A. (b) 12A. (c) 9A. click here
15) Using the formula for equivalent resistance of parallel resistors, the equivalent resistance is (a) 2.67Ω. (b) 26Ω. (c) 6Ω.
16) The current that leaves the battery is controlled by the equivalent resistance that the battery faces. Based on the equivalent resistance, the current that is pulled from the battery is (a) 9A. (b) 12A. (c) 24A. click here
Problem: Let a 36-volt battery be connected to following three resistors : R1 = 24Ω , R2 = 48Ω, and R3 = 14Ω. Let R1 and R2 be in parallel between points a and b, and place R3 between points b and c. In your calculations to follow, name the resistance between a and b as Rab, and the resistance between b and c, as Rbc. It is very important to draw a diagram for the problem. Answer the following questions: click here
17) Rab, the equivalent resistance for the two parallel resistors R1 and R2 is (a) 72Ω. (b) 24Ω. (c) 16Ω.
18) Rac, the total resistance the battery is facing is (a) 86Ω. (b) 30.Ω. (c) 10.4Ω.
19) If the total resistance that the 36-V battery is facing is 30Ω, the current that the battery is allowed to deliver is (a) 1.2A. (b) 2.4A. (c) 3.6A. click here
20) The current through Rac-portion is (a) 2.57A. (b) 1.2A. (c) equal to the current that flows out of the battery. (d) b & c.
21) The current through Rab-portion is (a) 2.25A. (b) 1.2A. (c) equal to the current that flows out of the battery. (d) equal to the current that flows through the Rac-portion. (e) b, c, & d. click here
22) To find the voltage (Potential difference) between points a and b, named "Vab", (a) Rab must be multiplied by the 1.2A. (b) just think that Vab is equal to the 36V that the battery supplies. (c) neither a nor b.
23) To find the voltage between b and c, or Vbc, (a) Rbc must be multiplied by the 1.2A. (b) just think that Vbc is also equal to the 36V that the battery supplies. (c) neither a nor b. click here
24) The current that flows through R1 is (a) more than 1.2A. (b) less than 1.2A. (c) equal to 1.2A.
25) The current in any portion of the circuit in this problem (a) can become more than 1.2A. (b) cannot be more than 1.2A. (c) must be set equal to 1.2A, even through R1 alone or R2 alone. click here
26) Since Vab is known as well as R1, the current I1 through R1 is (a) 19.2V / 24Ω = 0.80A. (b) 36V / 24Ω = 1.5A. (c) just 1.2A.
27) Since Vab is known as well as R2, the current I2 through R2 is (a) 19.2V / 48Ω = 0.40A. (b) 36V / 24Ω = 1.5A. (c) just 1.2A.
28) The sum of the currents in R1 and R2 must be (a) 2.4A. (b) 3.6A. (c) 1.2A. click here
29) The sum of the voltage drops across the ab-portion and the bc-portion must be (a) 72V. (b) 36V. (c) 0.0
30) The power dissipation in R1, R2 , and R3 are: (a) 15.36w, 7.68w, & 20.16w. (b) 24w, 48w, & 14w. (c) 20w, 40w, & 11.7w. click here
The Effect of Temperature on Resistance:
Resistivity ( ρ) increases with increase in temperature (T). This change is linear and given by
ρT = ρ20 [ 1 + α ( T - 20) ]
where ρT is the resistivity at temperature T, and ρ20 is the resistivity at 20º Celsius. The symbol ( α ) denotes the temperature coefficient of resistivity. Values of (α) are measured for different materials at room temperature (20ºC) and typical values are given in Table 20.1.
For a wire, when temperature increases, length-increase causes a resistance-increase while cross-section-increase causes resistance-decrease. Assuming these two effects to somehow offset each other, for tiny wires we may approximately write: R T = R20 [ 1 + α ( T - 20) ]. With alternating currents passing through light bulb elements, there are other effects that we neglect those here for simplicity.
Example 16: The resistance of a light-bulb is measured to be 18.2Ω when cold ( not when connected to wall 120-V outlet). However, the bulb is rated as a 60-watt light bulb ( This means that when hot and connected to a 120-V outlet it consumes 60 watts.) (a) Calculate the current it pulls and its resistance when hot. (b) What temperature does it have when hot? The elements of light bulbs are made of tungsten [α = 4.5x10-3 (º C)-1].
Solution: P = VI or, I = P / I = 60watts / 120V = 0.50 Amps. (when hot)
R = V / I = 120V / 0.50A = 240Ω ( Compare 18.2 Ω with 240 Ohms).
Using the approximate R T = R20 [ 1 + a ( T - 20) ], we get 240 = 18.2 [ 1 + 4.5x10-3 ( T - 20.) ] from which T = 2728 rounded to 2700 ºC. The melting point of tungsten is 3400 ºC.
Sometimes the electric elements are connected in such a way that it is not possible to label them as being in series or parallel. This happens when there are loops or closed circuits in the system. For example, in the figure shown below, we can not say that R1, R4, and R3 are connected in parallel or in series. We also see that there are two loops: Loop ABCDA and Loop ABFEA. Between A and B, there are 3 branches. There will be 3 different currents in these 3 branches. Elements R1, R2, and battery V1 experience same current. This current is labeled I1. Element R4, and battery V2 experience a different current. This current is labeled I2. The elements in the last branch (AGB), R3 and V3, experience another current that is named I3. The purpose is to calculate I1, I2, and I3. To do this, we use the so-called "Kirchhoff's Rules."
KLR in Loop ABCDA:
-V2 + I3R3 + V1 + I1R2 + I 1R1 = 0 (1)
KLR in Loop ABFEA:
-V2 + I3R3 - I2R4 + V3 = 0 (2)
KJR at Node A:
- I1 + I3 + I2 = 0 (3)
Kirchhoff's Loop Rule (KLR):
This rule states that the sum of voltage ups and downs in a closed loop is equal to zero. In each branch, we don't know the direction of the current. We simply assume a direction for it. If our assumption is correct, the answer we find at the end of our calculations will be positive. If our guess is wrong, the calculations will yield a negative value for that current. The negative sign indicates that we chose a wrong direction. To write down the ups and downs of voltage in a closed loop, we need to start from a point on the loop, go around it, complete it, and come back to that point. In two ways, a loop can be completed. 1) By going with the assumed direction for current, and 2) by going against the assumed direction for current.
If we follow the assumed direction for the current, it is like swimming with the flow of a river. It is swimming toward points of lower and lower potentials. This way going through a resistor will cause a drop in voltage (potential) and we give a negative sign for the amount of voltage drop. If we encounter a battery on our way, if its polarity is such that we go from (-) to (+), we consider a positive value for its voltage, and if we go from (+) to (-), we write down a negative value for its voltage.
Now, if the way we complete a closed loop is opposite to the assumed direction for the current, it is like swimming against the flow of a river or simply swimming toward points of higher elevations and potentials. This way when we encounter a resistor in the circuit, we write the potential increase across it as a positive value. For a battery, if we go through it by going from (+) to (-), the voltage change will be negative, and if from (-) to (+), the voltage change will be positive. A KLR is written for each loop of the above figure.
Kirchhoff's Junction Rule ( KJR ):
The sum of currents going toward a junction is equal to the sum of currents leaving that junction. In other words, the algebraic sum of currents at a junction is zero. A junction is the point of connection of 3 or more wires. A KJR is written at Junction A in the above figure.
Example 17: In the figure shown, the directions for currents I1, I2, and I3 are already chosen. Find their values and state whether the assumed directions are correct?
Solution: KLR in Loop ABCDA: + 40 I3 - 5 + 6 + 30 I1 + 20 I1 = 0
KLR in Loop ABFEA: + 40 I3 - 5 - 9 + 20 I2 + 50 I2 = 0
KJR at Node B : + I1 + I2 - I3 = 0
Making the above equations ready for calculator, we get:
+ 50 I1 + 0 I2 + 40 I3 = -1
0 I1 + 70 I2 + 40 I3 = 14
+ 1 I1 + 1 I2 - 1 I3 = 0
|Results of calculation:
I1 = - 0.0807 A
I2 = 0.1566 A
I3 = 0.0759 A
|In the above figure, the direction of I1 must be reversed.|
Circuits containing a resistor and a capacitor form the so-called RC-circuits. A typical RC circuit in charging process requires a battery as shown below. The resistance R indeed controls the speed of the charging process and depending on the capacity C of the capacitor, the time involved in the charging a capacitor varies. The product RC is called the time constant ( τ )of the circuit. τ is pronounced " tau ."
τ = RC has unit of time, in SI, seconds.
|Current as a function of time during the
charging process is an exponential function of time and is given by:
I = Imax e -(t / RC) where
I max = Vb / R
As soon as key K is closed the battery with voltage Vb faces an empty capacitor C and a resistor R. At the very beginning, the capacitor, being empty, does not show any resistance toward filling up. Therefore at the start of the process, the only element showing resistance is R, and that decides the amount of the initial current ( I0). I0 can also be shown as Imax . Maximum current occurs at the very beginning. As the capacitor fills up, its resistance goes up and the battery faces more resistance. This causes the current go down as the capacitor fills up. When the capacitor is fully charged, its voltage equals the battery voltage. At this stage the current through the circuit approaches zero and so does the voltage across the resistor. At any given instant, Vb = VR + VC. As VC increases, the current ( I ) goes down causing VR to decrease.
Example 18: In the figure shown, find Io ( Imax ), and I(1sec), I(5sec), I(10sec), and I(100sec). Assume 3 significant figures for all numbers.
|RC = (100Ω)(0.05 Farads) = 5.00
I max = V / R = 20 volts / 100Ω = 0.2 A
I ( t ) = Imax e ( -t / RC)
I ( 1 ) = 0.2 exp ( - 1 / 5 ) = 0.164 A
I ( 5 ) = 0.2 exp ( - 5 / 5 ) = 0.0736 A
I ( 10 ) = 0.2 exp ( - 10 / 5 ) = 0.0270A
I ( 100 ) = 0.2 exp ( - 100 / 5 ) = 0 A
Example 19: In the previous example, calculate VR and VC at the given instants.
|t = 0||VR = RI = (100 Ω)(0.200 A) = 20 volts||VC = Vb - VR = 20 - 20 = 0 volt|
|t = 1||VR = RI = (100 Ω)(0.164 A) = 16.4 volts||VC = Vb - VR = 20 - 16.4 = 3.6 volt|
|t = 5||VR = RI = (100 Ω)(0.0736 A) = 7.36 volts||VC = Vb - VR = 20 - 7.36 = 12.64 volt|
|t = 10||VR = RI = (100 Ω)(0.0270 A) = 2.70 volts||VC = Vb - VR = 20 - 2.70 = 17.30 volt|
|t = 100||VR = RI = (100 Ω)(0 A) = 0 volts||VC = Vb - VR = 20 - 0 = 20.0 volt|
Test Yourself 4:
1) The resistance of a resistor (a) increases with temperature increase. (b) decreases with temperature increase. (c) is not a function of temperature. click here
2) Knowing that ρ20 is the resistivity of a material at 20ºC, the temperature coefficient of electric resistivity, α is (a) the change in resistivity, Δρ. (b) the change in resistivity per unit resistivity, Δρ /ρ20. (c) the change in resistivity per unit resistivity per unit change in temperature, Δρ /(ρ20ΔT). click here
3) If we write α as α = Δρ /(ρ20ΔT) that you may want to write it with a horizontal fraction bar, it is easy to find out that the unit of α is (a) ºC-1. (b) Ω / ºC. (c) [Ω ºC]-1.
4) One of the elements used in the filament of light bulbs is tungsten. The resistivity of tungsten (From Table 20.1) at 20ºC is ρ20 = 5.6x10-8 Ωm and its temperature coefficient of resistivity is α = 4.5x10-3[ºC]-1. The resistivity of tungsten at 2720ºC is (a) ρ2720 ºC = 7.3x10-7 Ωm. (b) ρ2720 ºC = 5.9x10-7 / ºC. (c) neither a nor b. click here
Problem: A student uses an ohmmeter to measure the resistance of a light bulb when cold and not in use. Its cold resistance is R20 ºC = 11Ω. If the element is made of tungsten, and we neglect the changes in its dimensions due to temperature increase, find its resistance at 2700ºC. We may approximately use RT = R20[1 + α (T - 20) ] instead of ρT = ρ20[1 + α (T - 20) ]. Answer the following questions:
5) The resistance at 2700ºC of the element is (a) 129Ω. (b) 144Ω. (c) 192Ω. click here
6) Since light bulbs are designed to operate at 120V in the United States, the current this light bulb draws from a wall socket is (a) 1.2A. (b) 0.833A. (c) 17000A.
7) The power of the light bulb may be calculated by (a) P = VI. (b) P = RI2. (c) both a & b. click here
8) The power of the light bulb is (a) 150watts. (b) 75watts. (c) 100watts.
9) If you pick up a 100-watt GE light bulb and measure its resistance cold, you expect to measure a resistance of (a) 144Ω. (b) about 11Ω. (c) about 72Ω. click here
10) What current does a 60-watt light bulb draw from a wall socket when in use? (a) 0.5A (b) 2A (c) 1A.
11) What is the resistance of a 60-watt light bulb when in use (Assume 2700ºC.)? (a) 240Ω. (b) 120Ω. (c) 180Ω.
12) What is the resistance of a 60-watt light bulb when cold and not in use (assume 20ºC)? (a) 9Ω. (b) 5Ω. (c)18.4Ω.
13) According to Kirchhoff's Loop Rule (KLR), the sum of (a) battery voltages must be zero. (b) voltages across all elements must be zero. (c) voltage ups and downs across the elements in a closed loop is zero. click here
14) A junction is a point in an electric circuit to which (a) only two wires are connected. (b) 3 or more wires are connected. (c) only a single wire is connectd.
15) According to Kirchhoff's Junction Rule (KJR), the sum of (a) currents going to and from a junction is zero. (b) the sum of voltages at a junction is zero. (c) neither a nor b. click here
Problem: Draw a square loop, and starting from the lower left corner, move in the clockwise direction, and label it abcd. In the ab side draw a resistor and label it 10Ω. In the bc side place a 3-V battery with its pos. pole on the left. In the cd side place an 18-V battery with its neg. pole up and follow it by a 20-Ω resistor. In the da side place a 15-Ω resistor. Answer the following questions: click here
We want to write a KLR for loop abcda. We do not know the actual current direction. We may assume it clockwise or counterclockwise. Let's assume it clockwise. If we start from point b, for example, and go clockwise, we are going with the flow or current, I. click here
16) From b to c, we have to go through the 3-V battery from its pos. pole to its neg. pole. This is associated with a (a) voltage drop and we write -3 volts. (b) voltage increase and we write +3 volts. (c) neither increase nor decrease.
17) From c to d, we have to go through the 18-V battery from its neg. pole to its pos. pole. This is associated with a (a) voltage drop and we write -18 volts. (b) voltage increase and we write +18 volts. (c) neither increase nor decrease.
18) After the 18-V battery, there comes the 20-Ω resistor. Since we are following the assumed direction for the current, I, we are moving toward lower potentials. The voltage difference across this 20-Ω resistor must be taken to be negative and we write (a) -20I. (b) +20I. (c) +20. click here
19) From d to a , going with the flow is again associated with another (a) voltage increase across the 15-Ω resistor and we write +15I. (b) voltage decrease across the 15-Ω resistor and we write -15I.. (c) voltage decrease across the 15-Ω resistor and we write -15.
20) Going from a to b to complete the loop, we run into the 10-Ω resistor only. Again, going with the assumed direction for I, there is a (a) voltage increase across the 10-Ω resistor and we write +10I. (b) voltage decrease across the 10-Ω resistor and we write -10I.. (c) voltage decrease across the 10-Ω resistor and we write -10. click here
21) The completed KLR to Loop abcda is (a) -3+18-20I-15I-10I. (b) -3+18-20I-15I-10I.= 0 (c) neither a nor b.
22) Solving for I, the current in the loop is (a) 3.0A (b) 6.0A (c) 0.33A click here
23) Since the current turned out positive, our assumed clockwise direction is (a) correct. (b) wrong. (c) partially correct.
Now solve the problem assuming a counterclockwise direction for the unknown current I. Redraw the diagram with the new assumed direction for the current. However, start from b again and go clockwise again, that is opposite to the assumed direction for the current, I. click here
24) From b to c, we have to go through the 3-V battery from its pos. pole to its neg. pole. This is associated with a (a) voltage drop and we write -3 volts. (b) voltage increase and we write +3 volts. (c) neither increase nor decrease.
25) From c to d, we have to go through the 18-V battery from its neg. pole to its pos. pole. This is associated with a (a) voltage drop and we write -18 volts. (b) voltage increase and we write +18 volts. (c) neither increase nor decrease.
26) After the 18-V battery, there comes the 20-Ω resistor. Since we are going against the assumed direction for the current, I, we are moving toward higher potentials. The voltage difference across this 20-Ω resistor must be taken to be positive and we write (a) -20I. (b) +20I. (c) +20. click here
27) From d to a , going against the flow is again associated with another (a) voltage increase across the 15-Ω resistor and we write +15I. (b) voltage decrease across the 15-Ω resistor and we write -15I.. (c) voltage increase across the 15-Ω resistor and we write +15.
28) Going from a to b to complete the loop, we run into the 10-Ω resistor only. Again, going against the assumed direction for I, there is a (a) voltage increase across the 10-Ω resistor and we write +10I. (b) voltage decrease across the 10-Ω resistor and we write -10I.. (c) voltage increase across the 10-Ω resistor and we write +10. click here
29) The completed KLR to Loop abcda is (a) -3+18+20I+15I+10I. (b) -3+18+20I+15I+10I = 0. (c) neither a nor b.
30) Solving for I, the current in the loop is (a) -3.0A (b) -6.0A (c) -0.33A
31) Since the current turned out negative, our assumed direction is (a) correct. (b) wrong and must be reversed to clockwise. (c) partially correct. click here
Problem: Redraw Example 17, but double the battery voltages and triple the resistances. Also, reverse the assumed directions for I1 and I3. Answer the following questions:
32) KLR for ABCDA is (a) -120 -10 +12 -90I1 - 60I1 = 0. (b) -120I3 -10 +12 -90I1 - 60I1 = 0. (c) a & b.
33) KLR for ABFEA is (a) -120I3 -10 +18 -90I1 - 60I1 = 0. (b) -120I3 -10 -18 +60I2 +150I2 = 0. (c) a & b.
34) KJR at Node B is (a) I1 - I2 + I3 (b) I1 - I2 + I3 = 0 (c) - I1 + I2 + I3 = 0. click here
35) Answer for I1, I2, &I3 are: (a) 0.05382A, 0.10442A, -0.0506A (b) 53.82mA, 104.42mA, -50.60mA. (c) a&b.
Problem: Consider a RC circuit as shown in Fig. 9. Suppose that at t = 0 the capacitor is empty. As soon as key, K is turned on, the capacitor starts charging. Charges flow from the battery to the plates of the capacitor. As the capacitor accumulates charges on its plates, it builds up greater and greater voltages; however, the limiting voltage is the battery voltage that feeds it. Answer the following questions:
36) At t = 0, the voltage across the capacitor, VC is (a) 0, because there is no current in the circuit. (b) 0, because no charge is accumulated on its plates yet. (c) both a & b. click here
37) At t = 0, since VC = 0, all the battery voltage drop across (a) the resistance R. (b) the capacitor, C. (c) neither a nor b.
38) The initial current, Io, is Io = Vb /R because (a) the capacitor is initially full and does not accept any current; therefore, all the current flows toward the resistor. (b) the capacitor is empty and creates no resistance in the circuit. The only resistance to consume the battery voltage is R. (c) neither a nor b. click here
39) As the capacitor fills up it develops resistance toward accepting more charge and the total resistance in the circuit keeps going way beyond R. We may think that when the capacitor is nearly fully charged, the current in the circuit (a) approaches zero. (b) becomes very high. (c) neither a nor b.
40) As the current approaches zero, so does the (a) voltage across the capacitor. (b) the voltage across the resistor. (c) neither a nor b. click here
1) A 4.00-Amp current flows through a wire. Calculate the total electric charge flown in the wire in 50.0 minutes (a) in coulombs, and (b) in Ah.
2) If a car battery is put under charge for 24.00 hours at an average current of 1.500 Amps, calculate the stored charge in it (a) in Ah, and (b) in Coulombs.
3) A light bulb allows a current of 1.666 Amps when connected to a 120.-volt AC source. Calculate (a) its resistance, (b) the power it dissipates, and (c) its power dissipation if the voltage drops to 110.V.
4) Calculate (a) the resistance of a copper power line that is 375km long and 2.54cm thick (1.00 inch.). If this wire is passed through several dyes such that its diameter is reduced to half, (b) by what factor does its resistance increase? (c) Calculate its new resistance.
5) Calculate the resistance of a thin silver wire 0.0800mm thick and 60.0m long,
6) A 150-watt light bulb is designed for 220 Volts. (a) What current does it draw? If it is used for 5.0hours, how much energy does it consume (b) in Joules, and (c) in kwh? (d) If the cost of electric energy is 8.0¢/kwh, calculate the cost of keeping this bulb on for 30 full days. (e) What is its resistance when in use?
7) An electric water heater works with 240V and pulls 15Amps. Calculate its power in (a) watts, and (b) kilowatts. (c) For an energy cost of 8.0¢/kwh, calculate the monthly (30 days) expense of this heater for an average use of 5.0 hour per day.
8) An electric heater designed for 120.V has a cold resistance of 8.00Ω when at 20.°C. Find (a) the initial current it draws from the electric outlet as soon as it is turned on. If its normal operating temperature is 520.°C, and the temperature coefficient of resistivity of its element is 4.00x10-3[°C]-1, find (b) its resistance when hot. (c) What current does it draw when hot? Use the approximation: R(T)=R(20)[1+α(T-20)] instead of ρ(T)=ρ(20)[1+α(T-20)].
9) The temperature coefficient of resistivity of tungsten is α = 4.5x10-3/°C. Less efficient light bulbs were made of tungsten. At Room Temperature (20°C), the electric resistance of a 100-watt light bulb measures at R(20)=12.0Ω. You may calculate its hot resistance (when in use) by using its rated power of 100.w and rated voltage of 120.V. Calculate (a) its resistance when hot (in use). (b) Use the approximation:
R(T) = R(20) [ 1 + α ( T - 20 ) ], instead of ρ(T) = ρ(20) [ 1 + α ( T - 20 ) ], and find the temperature, T, of the light bulb when in use.
10) Resistors R1, R2, and R3 have values of 15Ω, 35Ω, and 25Ω, respectively and are in series with a 30.0V battery and an ammeter. Find the current in the circuit and the voltage across each resistor. Verify your solution by showing that the sum of the individual voltages you find equals the battery voltage.
11) If R1 and R2 in Prob. 10 are in a parallel module between points that you name A and B, and R3 is placed between points you name B and C, and the battery is between Points C and A, calculate (a) the voltage across each as well as (b) the current through each. Write your answers in order.
12) Draw two 4-inch horizontal parallel lines (like a railroad) 1 inch apart and divide each line into four 1-inch segments. With four segments on the top line you have 5 points. Name them A, B, C, D, and E. Name the points on the bottom line A', B', C', D' , and E'. Connect BB', CC', and DD'. Now, you should have a 3-step ladder. Suppose each segment is a 100.0-Ω resistor. If this module is connected to a battery at A and A', calculate the overall resistance between A and A'. Note that no current flows in segments DE and D'E' and those do not count.
13) Repeat problem 12 with points E and E' connected and of course EE' holds a 100.0-Ω resistor.
14) In a two-loop circuit, the left branch contains a 50.0-Ω resistor and a 9.50-V battery with its positive pole up. The middle branch has a 120.0-Ω resistor and a 4.00-V battery with its positive pole down. The right branch contains a 40.0-Ω resistor and a 10.8-V battery with its positive pole up. Naming the currents I1, I2, and I3 for the left, middle, and right branches, respectively, find the current in each branch and write them in the same order.
15) A 25.0μF capacitor is in series with a 120.0kΩ resistor and a 15.0-V battery. If the capacitor is initially empty, find the values of VC and IC at the following instants: 0.00s, 3.00s, 6.00s, 9.00s, 12.0s, and 15.0s. and write each in order.
1) 12,000Couloms, 3.33Ah 2) 36.00Ah, 129,600Coul.
3) 72.0 Ω, 200.watts, 168.watts 4) 12.6 Ω, 16, 202 Ω 5) 180Ω
6) 0.68A, 2.7x106J, 0.75kwh, $8.64, 320Ω 7) 3600w, 3.6kw, $43
8) 15.0A, 24.0Ω, 5.00A 9) 144Ω, 2464°C ≈ 2500°C
10) 0.40A, (6.0V, 14V, and 10.V)
11) (8.9, 8.9, & 21.1Volts), (0.593, 0.254, & 0.845A)
12) 273.3Ω 13) 273.2Ω 14) 30.0mA, 100.mA, & 70.0mA
15) (0, 9.48, 12.7, 14.3, 14.7, & 14.9V), (125, 46.0, 16.7, 6.22, 2.29, & 0.842μA)