Chapters 27 and 28
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, dq per change in time, dt 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: For a constant current, instead of the differential form I = dq /dt, we may write: I = Δq /Δt . Cross-multiplication solves for Δq. We get: Δq =IΔt or,
Δq = (5.00Amps)(20.0*60sec.) = 6000 Coul.
Besides Coulombs, another unit for the amount of electricity (q) is Ampere-hour (Ah). It is easy to see that:
1 Ah = 3600 Coul. The reason is : 1Ah = 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 ratio of voltageV across the device to the current I through the device is a constant. The 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 any cross-section of a wire, 480 Coulombs of electric charge pass per minute, the currents is (a) 8.0 Amps (b) 480 Amps (c) 480 mA.
4) If through any cross-section of a wire, 7200 μCoul. of electric charge pass every hour, the current is (a) 2.0A (b) 2.0μA (c)2.0mA. click here
5) If through any cross-section of a wire, 108 μCoul. of electric charge pass 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) 20Ah (b) 36Ah (c) neither a nor b.
11) Ohm's Law states that (a) the voltage-to-current ratio for a resistor remains constant (b) the voltage-to-current ratio for 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) 20A (b) 0.02A (c) 26.4A.
29) If the resistance is changed 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 that same factor (b) increases by that same factor (c) remains unchanged. click here
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 is and the more forceful must the pushing of water through the pipe be. An electric wire acts in a similar manner. The greater its length, L, the smaller its cross-sectional area, A, and the greater its resistivity, ρ, the higher its electric resistance, R, is and it takes a greater voltage to push a higher current 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 20�C. Table 20.1 gives the values of ρ for a few selected materials:
|ρ ( Ωm )||α ( �C ) -1||ρ ( Ωm )||α ( �C ) -1|
Aluminum 2.82 x 10-8
Copper 1.70 x 10-8
Iron 10 x 10-8
Nickel 7.8 x 10-8
Platinum 10 x 10-8
Tungsten 5.6 x 10-8
4.29 x 10-3
6.80 x 10-3
6.51 x 10-3
6.0 x 10-3
3.93 x 10-3
4.5 x 10-3
Carbon 3.6 x 10-5
Germanium 4.6 x 10-1
Silicon 2.5 x 10 2
-5.0 x 10-2
-7.0 x 10-2
Example 4: Calculate the resistance of a copper power line that is 425km long and 1.27cm thick (1/2 inch.).
Solution: The diameter of the circular wire is D =1.27cm. 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.
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 (5.00x10-5)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 1watt = ( 1volt )( 1amp ).
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 = (86,400,000J) / (4.186 J/cal) = 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 the energy or work, in kwh.
Example 9: A water heater works at 240V 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 energy consumption in kwh. If electric energy sells for 9.00 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 )( $ 0.0900 / kwh ) = $ 97.2.
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 thickness.
5) Two wires A and B have the same length and the same material. The diameter of wire A is 1/3 of that of wire 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. 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 vacuum cleaner is 120 volts and the current through it is 4.0A, the power dissipation in it is (a) 480 watts (b) 90. watts (c) 560watts.
22) The voltage across a vacuum cleaner 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) A vacuum cleaner with an in-use resistance of 30.Ω allows a 4.0-A current to flow through it. Its power 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 rated as 75 watts is designed for 120V. 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)( 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 75.0watt 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 the 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 can push out 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 the1.2A (b) we think that Vab is just 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) we 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 20� Celsius. The symbol ( α ) denotes the temperature coefficient of resistivity. Values of (α) measured for different materials at room temperature (20�C) are given in Table 20.1.
For a wire, temperature increases causes thermal expansion. Length increase causes resistance to increase while cross-section increase causes resistance to decrease. Assuming these two effects to somehow offset each other, for tiny wires we may approximately write: R T = R20 [ 1 + α ( T - 20) ]. When alternating currents pass through the elements of light bulbs, there are other effects that can affect the overall resistance, we neglect those here for simplicity.
Example 16: The resistance of a 60-watt light-bulb is measured 18.2Ω when cold (unlit). When connected to 120-V and is lit, calculate (a) the current it pulls as well as its resistance when hot. (b)What temperature does it have when hot? The elements of light bulbs are made of tungsten for which [α =4.5x10-3(�C)-1].
Solution: P = VI or, I = P / V = 60watts / 120V = 0.50 Amps. (when hot)
R = V / I = 120V / 0.50A = 240Ω ( Resistance when hot!).
Using the approximate formula 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 �C. The melting point of tungsten is 3400 �C.
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 jumps and drops in a closed loop is equal to zero. In each branch, we don't know the direction of the current to begin with. 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 jumps and drops of voltage in a closed loop, we need to start from a point of the loop, go around it, complete it, and come back to that point. In two ways, a loop can be traveled around. 1) By going with the assumed direction of current, and 2) by going against the assumed direction of current.
In adding voltages across the elements of any closed loop to set the total equal to zero,
If we follow the assumed direction for the current, it is like swimming with the flow of a river. Swimming with the current means going toward points of lower and lower potential. On this basis, traveling through a resistor with the assumed current direction means a drop in voltage and we give a negative sign to the product (RI) in the equation. (-RI) is the voltage drop across R.
Now, if we travel opposite to the assumed direction of current, it is like swimming against the flow of a river or simply swimming toward points of higher potential. On this basis, traveling through a resistor against the assumed current direction means a jump in voltage and we give a positive sign to to the product (RI) in the equation. (+RI) is the voltage jump across R.
It doesn't matter if we are traveling along a loop with the assumed current or against it, going from (-) to (+) is always a voltage jump (+Vbat) and going from (+) to (-) is always a voltage drop (-Vbat).
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 to and away from 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 determine if 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.|
A circuit containing a resistor and a capacitor forms the so-called "RC circuit." A typical RC circuit in which the capacitor is to be charged requires a battery as shown below. The resistance R is used to control the charging speed and depending on the capacity C, the charging time varies. The product RC is called the time constant ( τ )of the circuit. τ is pronounced " tau ."
τ = RC has unit of time, that is of course, seconds in SI.
Note: Imax , Io , and I(0) mean the same thing.
|The current in the circuit that is also the
current through the capacitor decreases as an exponential function of time and is given by:
IC = Imax e -(t / RC) where Imax = Vb/R
The capacitor voltage, VC, increases exponentially to its limiting value, Vbat, according to the following equation:
VC = Vmax[1 - e -(t / RC)] where Vmax = Vbat.
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 impose any resistance toward filling up. At t = 0, the only element showing resistance is R, and that determines the amount of the initial current ( Io). Io can also be shown as Imax . Maximum current occurs at the very beginning, t = 0. As the capacitor fills up, its develops resistance, and therefore the battery faces an increasing resistance in excess of R. This causes the current decrease 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, Vbat. = VR + VC. As VC increases, the current ( I or IC ) goes down causing VR to decrease and approach 0.
Example 18: In the figure shown, find (a) Io or Imax , and (b) the values of I at t = 1s, 5s, 10s, and 100s; in other words, find I(1), I(5), I(10), and I(100). Assume 3 significant figures on all numbers.
|RC = (100Ω)(0.05 Farads) = 5.00
I max = Vbat. / 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|
Example 20: In the previous example, find the values of Vc directly by using the equation for Vc, that is Vc = Vbat[ 1 - e (-t / RC) ] and compare the results with the values of VC in Example 19.
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 20�C, 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) �C-1 (b) Ω / �C (c) [Ω �C]-1.
4) One of the elements used in the filament of light bulbs is tungsten. The resistivity of tungsten (From Table 20.1) at 20�C is ρ20 = 5.6x10-8 Ωm and its temperature coefficient of resistivity is α = 4.5x10-3[�C]-1. The resistivity of tungsten at 2720�C is (a) ρ2720 �C = 7.3x10-7 Ωm (b) ρ2720 �C = 5.9x10-7 / �C (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 �C = 11Ω. If the element is made of tungsten, and we neglect the changes in its dimensions due to temperature increase, find its resistance at 2700�C. 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 2700�C 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 a filament temperature of 2700�C.) (a) 240Ω (b)120Ω (c)180Ω.
12) What is the resistance of a 60-watt light bulb when cold and not in use (assume 20�C)? (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 jumps and drops 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 connected.
15) According to Kirchhoff's Junction Rule (KJR), the sum of (a) currents going to and away 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 jump 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 voltage of the battery 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 = Vbat / 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, and the only resistance the battery faces 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 increasing 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
Problem: The derivation of RC Circuit Formula
|In the figure shown,
41) Write the voltage balance equation for Vbat, VR and VC that is good for any instant. See (*) below the figure.
42) In your answer for (41), Replace VR by its equivalent in terms of I. Also replace VC by its equivalent in terms of C and QC. What formula do you know that relates QC to C and VC? How does your new equation read now?
43) Do you know a differential form for the current I in terms of Q? Use it to write your result in (42) completely in terms if Q or indeed Q(t). This gives you the differential equation from which you can solve for Q(t).
44) One way of solving the differential equation you obtained in (43) is to multiply both sides of it by e^(t /RC). Do so and show that one side of your equation is exactly the time derivative of Qe^(t /RC). Continue and show that the final result after integration can be written as VC = Vbat[ 1 - e^(-t/RC)].
(*) Note that
Q = Q(t), I = I(t) , VC = VC(t) , and VR = VR(t).