Chapter 20
The Electric Current:
The current I through a wire is defined as the amount of charge q flown per unit of time t at every crosssectional area of the wire. Mathematically, we may write it as the time rate of change of electric charge.
The SI unit for I is of course Coul./sec called "Ampere", or Amp or simply A.
Example 1: A 5.00Amp current flows through a wire for 20.0 minutes. Calculate the amount of electric charges flown through the wire.
Solution: I = Δq /Δt. Solving for Δq, we get: Δq = IΔt or,
Δq = (5.00Amps)(20.0*60sec.) = 6000 Coul.
The Industrial Unit for Electric Charge:
The industrial unit for electric charge q is the "Amperehour (Ah)." A current of 1 Amp delivers 1Ah of charge if flown for 1 hour.
It is easy to see that 1Ah = 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 = 0.80A(40.0 h) = 32.0Ah
Δq = (32.0)(3600) Coul. = 115,000 Coul.
Ohm's Law:
For any linear electric device, the ratio of the voltage V across the device to the current I through the device is a constant called the "electric resistance, R" of the device. Mathematically,
where in SI, the unit of R is volts/amp called "Ohms" with the symbol Ω , pronounced "Omega." This means that if the voltage across an electric resistance (such as a light bulb) is doubled, the current through it also doubles to where the V/I ratio remains constant. Of course, when the current doubles, the excess heat it generates can burn the light bulb.
In electric circuits, batteries provide voltage (electric pressure). If a house lacks the city water supply, it then must have a water tank to generate mechanical pressure. The higher the water tank, the greater the pressure it generates that can be experienced when taking a shower. A higher water tank is exactly like a higher voltage battery placed in an electric circuit. If the elevation of a water tank is doubled, you can fill a pot twice as fast because the water current (# of gallons per minute) doubles. In an electric circuit, if you replace a battery of voltage V with one that has a voltage of 2xV, the current will also doubles from I to 2xI.
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 12V 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 selfexplanatory:
Solution: (Write with horiz. fraction bars). R = V/I = 12V/2.5A R = 4.8 Ohms, or R = 4.8 Ω When key K is closed, the circuit is turned on and current I flows turning the light bulb, R on. 

Figure 1
Test Yourself 1:
1) The electric current, I, is defined as (a) the # of gallons of water crossing every crosssection of a pipe per unit of time (b) the # of Coulombs of electric charge crosscrossing every section of a wire 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 crosssection of a wire, 480C of electric charge pass per minute, the currents is (a) 8.0 Amps (b) 480Amps (c) 480 mA.
4) If through any crosssection of a wire, 7200μC of electric charge pass every hour, the current is (a) 2.0A (b) 2.0μA (c)2.0mA. click here.
5) If through any crosssection of a wire, 108μC 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) 1.0Ah (b) 10Ah (c) 0.10Ah.
9) If a current of 5.0A passes through an electric device for 6.0h, the amount of electric charge Q flown through that device is (a) 30Ah (b) 108,000C (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 voltagetocurrent ratio for a resistor remains constant (b) the voltagetocurrent ratio for a resistor is not a constant (c) the coulombstocurrent 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 2.0Ωresistor is 3.0A. The voltage across it is (a) 6.0V (b) 0.67V (c) 1.5V.
16) The voltage across a 4.5Ωresistor 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 V_{bat.} = 12.0V, the voltage across the resistor, R (a) is 12.0V because there is only one resistor in the circuit and the entire battery voltage (effort) drops across that single resistor (b) would be 12.0V anyway even if other resistors R_{1}, R_{2}, R_{3}, ... 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 12V battery is connected to a 2.4Ω 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 2.4Ω 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 2.4Ω resistor is replaced by a 0.80Ω one (the resistance is 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 3.0A current is passing through an 8.0Ωresistor, the voltage across it is (a) 2.67V (b) 0.375V (c) 24V.
26) If the 8.0Ωresistance is doubled to 16Ω, the voltage across it (a) doubles to 48.0V (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.
The 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 crosssectional area, and the rougher it is inside, the more difficult it is to run water through it, and the more forceful must the pushing of water through the pipe be. An electric wire acts in a similar manner for the motion of electric current in it. The greater its length L, the smaller its crosssectional area A, and the greater its resistivity ρ, the higher its electric resistance R is and it takes a greater voltage to push a certain 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 of the wire. For copper, ρ_{cu} = 1.70x10^{8}Ωm at 20^{o}C. Table 20.1 gives the values of ρ for a few selected materials:
Table 20.1
ρ ( Ωm )  α ( ^{o}C ) ^{1}  ρ ( Ωm )  α ( ^{o}C ) ^{1} 
Conductors Aluminum 2.82x10^{8} Copper 1.70x10^{8} Iron 10x10^{8} Nickel 7.8x10^{8} Platinum 10x10^{}^{8} Tungsten 5.6x10^{8} 
4.29x10^{3} 6.80x10^{3} 6.51x10^{3} 6.0x10^{3} 3.93x10^{3} 4.5x10^{3} 
Semiconductors Carbon 3.6x10^{5} Germanium 4.6x10^{}^{1} Silicon 2.5x10^{2} Insulators Glass 10^{12} Rubber 10^{15} Wood 10^{10} 
5.0x10^{4} 5.0x10^{2} 7.0x 0^{2}

Example 4: Calculate the resistance of a copper power line that is 425km long and 2.54cm thick (1.00 inch.).
Solution: The diameter of the circular wire is D =2.54cm. R = 1.27cm = 1.27x10^{2}m.
R = ρL/A ; R = 1.7x10^{8} Ωm (425000m)/[π (1.27x10^{2})^{2}m^{2}] = 14.3Ω.
Example 5: Calculate the resistance of a thin copper wire 0.100mm thick and 50.0m long.
Solution: Since D =0.100mm, R = 0.0500mm = 5.00x10^{5}m.
R = ρL/A ; R = 1.7x10^{8} Ωm (50.0m)/[π (5.00x10^{5})^{2}m^{2}] = 108 Ω.
Power dissipation in a resistor:
From previous chapter, we had U_{e} = Vq. This may be written as
ΔU_{e} = VΔq. This means that when Δq charge flows at voltage V, it transfers, consumes, or delivers ΔU_{e} amount of energy.
An electric device that draws current I from a source at voltage V, consumes some electric energy U_{e} 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. Here is how:
The SI unit for power is J/s. Joule/sec is called "watt" as was discussed in Physics I. This means that 1watt = 1volt x 1amp.
Example 6: A lightbulb that works with 120V draws a current of 0.83A from a wall electric outlet. Find its power.
Solution: P = VI ; P = (120V)(0.83A) = 100 watts.
Example 7: A 60.0watt lightbulb is connected to a wall outlet. Calculate (a) the current it pulls and (b) its resistance.
Solution: (a) P = VI ; therefore, I = P/V; I = 60.0w/120V = 0.50A.
(b) V = RI ; R = V/I ; R =120V/0.5A = 240 Ω.
Other Versions of P = VI
Since V = R I, We may write P = (R I)( I ) or
P = R I^{2}. 
Also, from V = RI, solving for I, we get: I = V/R.
P = VI becomes: P = V (V/R), or
P = V^{2}/R. 
Example 8: A current of 5.00Amps flows through an electric heater with a resistance of 40.0Ω. Find (a) its electric power in watts, (b) its daily (24h) energy consumption in Joules, and (c) the daily heat energy in calories that it delivers. 1cal = 4.186J.
Solution: (a) P = RI^{2} = 40.0Ω(5.00A)^{2} = 1000 watts.
(b) P = U_{e} /t ; U_{e} = P t ; U_{e} = (1000 J/s)(24 x 3600s).
U_{e}= 86,400,000J. (c) U_{e} = 86,400,000J/(4.186 J/cal) = 2.06x10^{7}cal.
The Industrial Unit of Energy (kwh):
It is easy to show that kilowatthour (kwh) is a unit of energy. Work or energy U_{e} can be solved for in terms of power P and time t by crossmultiplying the formula P = U_{e } /t. This crossmultiplication yields:
U_{e} = P t ; If P is in kilowatts and t in hours, then U_{e} will be in kwh.
Example 9: A water heater that works at 240V pulls 12.5A. Find (a) its power in watts and kilowatts. If this heater is on 12.0h a day during a 30day month, find (b) its energy consumption in kwh. If electric energy sells for 11.0 cents per kwh, calculate (c) the cost per month of this heater.
Solution: (a) P = VI ; P = 240.V(12.5A) = 3000 watts ; P = 3.00kw.
(b) U_{e} = P t ; U_{e} = 3.00kw(360h) = 1080kwh.
(c) Cost = 1080kwh ( $0.110/kwh ) = $119.
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 crosssectional area A (d) directly on the reciprocal of its crosssectional area 1/A (e) a, b, & d. click here.
2) The resistivity ρ of a resistor depends on (a) the material of the resistor (b) the length of it (c) the crosssectional area of it (d) a, b, and c.
3) Comparing two wires of different lengths, but 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, but with different diameters, (a) the thicker wire has a greater resistance (b) the thicker wire has a smaller resistance (c) they both have the same resistance regardless of their diameters.
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) R_{A} = (1/3)R_{B} (b) R_{A} = 3R_{B} (c) R_{A} = 9R_{B}. click here.
6) For 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) R_{C} = 36R_{D} (b) R_{C} = 6R_{D} (c) R_{C} = (1/6)R_{D}.
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 (135ft) 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 it is 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 an orifice 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 crosssectional area is reduced by a factor of 4 (b) A = π r^{2} (c) the volume of the wire has to remain constant in the process. If the crosssectional area is reduced by a factor of 4, the length has to increase by a factor of 4 for the volume to be constant. (d) a, b, & c.
12) If a wire is passed through an orifice 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 an orifice to where its crosssectional 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 an orifice 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 ρ of metals is of the order of (a)10^{8}Ωm (b)10^{5}Ωm (c)10^{+12}Ωm.
16) The ρ of germanium is of the order of (a)10^{8}Ωm (b)10^{1}Ωm (c) 10^{+12} Ωm.
17) The ρ of 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 = RI^{2} (d) a, b, & c.
20) If the voltage across a resistor is 15V and the current through it is 2.0A, the power dissipation in it is (a) 7.5watts (b) 30w (c) 60w. click here.
21) The voltage across a vacuum cleaner is 120V and the current through it is 4.0A. Its power consumption is (a) 480w (b) 90 w (c) 560w.
22) The voltage across a vacuum cleaner is 120 volts and it pulls 4.0A, its electric resistance (while in use) is (a) 480Ω (b) 160Ω (c) 30Ω. click here.
23) A vacuum cleaner with an inuse resistance of 30Ω allows a 4.0A current to flow through it. Its power is (a) 30(4.0)^{2}w (b) 480w (c) both a & b.
24) One version of the power formula P = VI is (a) P = RI^{2} (b) P = RI (c) both a & b. click here.
25) An electric iron works with 120V and its inuse electric resistance is 15Ω. The electric current it pulls from the socket and the power it dissipates as heat are (a)10A &1000w (b)8.0A & 960w (c)15A & 1800w.
26) A light bulb rated as 75 watts is designed for 120V. The current it pulls from the socket and its inuse resistance are (a) 0.625A & 192Ω (b)1.25A & 384Ω (c) 1.60A & 75Ω. click here.
27) P = VI may be written as (a)P = V(q/t) (b)P = Vq/t (c)P = U_{e}/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) U_{e} = P t (c) both a & b.
29) The power of a light bulb is 150w. The electric energy U_{e} it consumes in 30s is (a) 5.0J (b) 4500J (c) 0.20J.
30) An electric motor is rated as 2.50hp. Its power in watts is (a) 1890w (b) 1380w (c) 790w.
31) If a 2.50hp 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 (kilowatt) of a 2300w electric motor is (a) 2,300,000kw (b)2.3kw (c) neither a nor b.
33) The power in kw of a 150w lightbulb is (a) 0.15kw (b)150,000kw (c) neither a nor b. click here.
34) The power in kw of a 4.00hp electric pump is (a)4.00kw (b)6.00kw (c)2.98kw. click here.
35) The electric energy U_{e} a 4.00hp pump consumes in 2.0h expressed in kwh is (a) 5.96kwh (b)12.0kwh (c)18kwh.
36) A 100w 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)1.04x10^{8}J, 28.8kwh (c) neither a nor b. click here.
38) For a 13.0 cents per kwh energy price, the cost of leaving a 75.0watt lightbulb on for 5.00 days is (a) $1.77 (b) $1.50 (c) $1.17.
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, R_{2}, and R_{3} are in series, there is a single resistor R_{t} (called the total or the equivalent resistance) that can replace them all such that
R_{t} = R_{1} + R_{2} + R_{3}
In a series circuit as shown in Fig. 2, below:1) The current I is the same in all resistors, but the voltagedrop across each resistor is different.
2) If R_{1} existed only, the battery voltage V_{bat} would be totally dropped across R_{1}. Now with R_{2} and R_{3} present, V_{bat} will be divided proportionally amongst the 3 resistors such that:
V_{bat} = V_{1} + V_{2} + V_{3}. Applying Ohm's law to individual resistors, results in: V_{1}= R_{1}I , V_{2}= R_{2}I and V_{3}= R_{3}I ; Consequently,
V_{bat} = R_{1}I + R_{2}I + R_{3}I. V_{bat} is also the voltage across the desired resistance R_{t} that is supposed to replace R_{1},R_{2}, and R_{3}. Ohm's law requires that V_{bat} = R_{t} I. Now, V_{bat} = V_{1} + V_{2} + V_{3} becomes:
R_{t} I = R_{1}I + R_{2}I + R_{3}I. simplifying:
R_{t} = R_{1} + R_{2} + R_{3.}

Figure 2
Example 10: In the above figure, let V_{bat} = 20.0 volts, R_{1} = 25.0Ω, R_{2} = 50.0Ω, and R_{3} = 25.0Ω. Calculate R_{t}, I, V_{1}, V_{2}, and V_{3}.
Solution: R_{t} = R_{1}+R_{2}+R_{3} = 100.0Ω.
Using Ohm's Law: I = V_{bat}/R_{t} = 20.0V/100.0Ω = 0.200A.
Now: V_{1} = R_{1}I = 25.0Ω(0.200A) = 5.00V.
V_{2 }= R_{2}I = 50.0Ω(0.200A) = 10.0V.
V_{3} = R_{3}I = 25.0Ω(0.200A) = 5.00V.
Resistors in Parallel:
Parallel Resistors experience the same voltage but different currents. Resistors that are between the same two nodes experience the same voltage and form a parallel module (Fig. 3). At one node the arriving current divides amongst the resistors and at the other node they join to form the main current again. The greater a resistor, the smaller the current its branch draws. Parallel resistors divide the main current amongst themselves proportional to 1/R. The voltage for all resistors between two nodes (in a parallel module) is the same. The equivalent resistance R_{eq} may be calculated from the formula:
In Fig. 3, 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 V_{dc} and V_{ab}, we get the same reading. V_{dc} = V_{ab }= V_{bat} . Since V_{ab} is the same as V_{bat} , we may calculate the currents in each branch by using the Ohm's law:
I_{1} = V_{ab}/R_{1}, similarly: I_{2} = V_{ab}/R_{2} and I_{3} = V_{ab}/R_{3}.
The 3 resistors can be replaced by R_{eq} such that I = V_{bat }/R_{eq}.
I = I_{1} + I_{2} + I_{3 } or V_{bat }/R_{eq}= V_{bat} /R_{1} + V_{bat} /R_{2} + V_{bat} /R_{3}_{ } .
Dividing by V_{bat}, we get: 1/R_{eq} = 1/R_{1} + 1/R_{2} + 1/R_{3}

Figure 3
Example 11: In Fig. 4, find the equivalent resistance in each case.
Figure 4
Solution: In Part 1, the two resistors are in series and they simply add up to a greater resistance (15Ω).
In Part 2, the two resistors are in parallel and their reciprocals add up to yield the reciprocal of the equivalent that is smaller than the smaller resistor (6Ω ).
In Part 3, the two parallel resistors yield 8Ω. The 8Ω is then in series with the 9Ω . They add up to 17Ω.
In Part 4, each parallel module must be calculated first. The first module yields 18Ω, and the one 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: To be done by students.
Example 13: In Fig. 3, let R_{1} = 30.0Ω, R_{2} = 6.00Ω, and R_{3} = 3.33Ω.
Also let V_{bat} = 4.00 volts. Find R_{eq}, I, I_{1}, I_{2}, and I_{3}.
Solution: R_{eq} = 1/(1/R_{1} +1/R_{2} +1/R_{3} ) = 2.00Ω. Ohm's Law:
I = V_{bat} / R_{eq} ; I = 4 volts / 2Ω = 2.00A.
V_{ab} = V_{dc} ;therefore, I_{1} = V_{ab }/R_{1} = 4V/30Ω = 0.133A.
I_{2} = V_{ab }/ R_{2} = 4V/6Ω = 0.667A.
I_{3} = V_{ab }/ R_{3} = 4V/3.33Ω = 1.20A.
Obviously : I_{1} + I_{2} + I_{3} = 2.00A. (Verify).
Mixed Series and Parallel Circuits:
There are many cases in which a parallel module is in series with other modules. The other modules may also be comprised of series or parallel modules themselves. 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.
R_{1} = 12 Ω, R_{2} = 60. Ω, R_{3} = 5.0 Ω, and V_{bat} = 45 volts.
Solution:
R_{ab} = 1/(1/ R_{1} + 1/ R_{2}) = 10Ω. R_{ac} = R_{ab}+ R_{bc} = 10 + 5.0 = 15Ω. I = V_{bat}/R_{ac} = 45volts/15Ω = 3.0A. V_{bat }partially drops across R_{3} that is V_{bc} and partially across the parallel module V_{ab}. V_{bc} = R_{bc}I = 5.0Ω(3.0A) = 15V. V_{ab} = V_{ac} V_{bc} = 45  15 = 30V. I_{1} = V_{ab}/R_{1} = 30V/12Ω = 2.5A. I_{2} = V_{ab}/R_{2} = 30V/60Ω = 0.50A. P_{1} = V_{1}I_{1} = 30V(2.5A) = 75 watts. P_{2} = 30V(0.50A) = 15 watts. P_{3} = 15V(3.0A) = 45 watts.

The voltmeter across the parallel module ab reads V_{1} and V_{2} that are the same. The voltmeter across bc reads V_{3} that is the voltage across R_{3}.

Figure 5
Example 15: In the figure shown, find the equivalent resistance between points a and b.
Figure 6
Solution: In the Problem, if a current flows from a to m, it has to divide at m, partially toward n, and partially toward s. The part that arrives at n must also divide, partially flow through the 5Ωresistor and partially through the 60Ωresistor. Looking at the top left figure, the part that flows through the 5Ωresistor must also flow through the 15Ω resistor as well; therefore, the 5Ω and 15Ωresistors are in series. The first step is to simply add the 5Ω and 15Ωresistors to get an equivalent of 20Ω resistance as shown in Part 1. In Part 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 Part 2.
In Part 2, the 25Ω and 15Ωresistors are in series because whatever current flows through the 25Ω must also flow through the 15Ωresistor. They add up to 40Ω as shown in Part 3. In Part 3 , the 40Ω and the 120Ω resistors are in parallel because the current that arrive at m is a must divide. They can be replaced by their equivalent resistance of (120*40)/(120+40) = 30Ω as shown in Part 4. In Part 4, the 50Ω and the 30Ωresistors are in series because the current through the 50Ωresistor must also flow through the 30Ωresistor. 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) In general, resistors in mixed series and parallel circuits experience (a) the same current (b) the same voltage (c) neither a nor b.
4) The equivalent 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) The equivalent 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 R_{eq} for a 120Ω, an 80Ω, and a 48Ω resistor connected in series is (a) 96Ω (b) 248Ω (c) 280Ω.
7) The R_{eq} for a 120Ω, an 80Ω, and a 48Ω resistor connected in parallel is (a) 96Ω (b) 248Ω (c) 24Ω.
8) R_{1} = 120Ω and R_{2} = 80Ω are in parallel and the result is in series with R_{3} = 48Ω. Draw a figure and place the parallel set between points A and B, and the third resistor, R_{3}, 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 module between E and F is in series with the module 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 value resistors and triple the odd value ones. 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, R_{1} = 50Ω, R_{2} = 150Ω, and R_{3} = 100Ω. Redraw the figure. Without looking at the solution, recalculate new values. The values of R_{t }, I , V_{1 }, V_{2 }, and V_{3} 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. Assume 3 sig. figs. on all values.
Problem: Let a 24volt battery be connected to following three resistors in parallel : R_{1} = 8Ω , R_{2} = 12Ω, and R_{3} = 6Ω. Draw a diagram for the problem. Answer the following questions: click here.
12) We may say that the voltage across (a) R_{1 }is 24volts (b) R_{2 }is 24volts (c) R_{3 }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 36volt battery be connected to the following three resistors: R_{1} = 24Ω, R_{2} = 48Ω, and R_{3} =14Ω. Let R_{1} and R_{2 }be in parallel between points a and b, and place R_{3 }between points b and c. In your calculations to follow, name the resistance between a and b as R_{ab}, and the resistance between b and c, as R_{bc}. It is very important to draw a diagram for the problem. Answer the following questions: click here.
17) R_{ab}, the equivalent resistance for the two parallel resistors R_{1} and R_{2 }is (a)72Ω (b)24Ω (c)16Ω.
18) R_{ac}, the overall resistance the battery is facing is (a)86Ω (b)30.Ω (c)10.4Ω.
19) If the total resistance that the 36V battery is facing is 30Ω, the current that the battery can push out into the circuit is (a)1.2A (b)2.4A (c)3.6A. click here.
20) The current through R_{ac } 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 R_{ab} 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 R_{ac} (e) b, c, & d. click here.
22) To find the voltage between points a and b, or V_{ab} , (a) R_{ab }must be multiplied by 1.2A (b) we think that V_{ab} is just equal to the battery voltage (c) neither a nor b.
23) To find the voltage between b and c, or V_{bc}, (a) R_{bc }must be multiplied by the 1.2A (b) we think that V_{bc} is just equal to the battery voltage (c) neither a nor b. click here.
24) The current that flows through R_{1} is (a) more than 1.2A (b) less than 1.2A (c) equal to 1.2A.
25) The current in any portion of this circuit (a) can be more than 1.2A (b) cannot be more than 1.2A (c) must be set equal to 1.2A, even in the absence one of R_{1} or R_{2}. click here.
26) Since V_{ab} is known as well as R_{1}, the current I_{1} through R_{1} is (a) 19.2V/24Ω = 0.80A (b) 36V/24Ω = 1.5A (c) just 1.2A.
27) Since V_{ab} is known as well as R_{2}, the current I_{2} through R_{2} is (a) 19.2V/48Ω = 0.40A (b) 36V/24Ω = 1.5A (c) just 1.2A.
28) The sum of the currents in R_{1 }and R_{2 }must be (a)2.4A (b)3.6A (c)1.2A. click here.
29) The sum of the voltage drops across the abportion and the bcportion must be (a) 72V (b) 36V (c) 0.
30) The power dissipation in R_{1}, R_{2 }, and R_{3 }are: (a) 15.36w, 7.68w, & 20.16w (b) 24w, 48w, & 14w (c) 20w, 40w, & 11.7w. click here.
The Effect of Temperature on Resistivity:
Resistivity ρ increases with increase in temperature T. This change is linear and given by
ρ_{T} = ρ_{20 }[ 1 + α ( T  20) ]
where ρ_{T }is the resistivity at temperature T, and ρ_{20 }is the resistivity at 20^{o} Celsius. The symbol α denotes the "temperature coefficient of resistivity." Values of α measured for different materials at room temperature (20^{o}C) are given in Table 20.1.
For a wire, any temperature increase causes thermal expansion. Length increase causes resistance to increase while crosssection increase causes resistance to decrease. Assuming these two effects to somehow offset each other, for tiny wires we may approximately write:
R _{T} = R_{20 }[ 1 + α ( T  20) ].
When alternating currents pass through the elements of light bulbs, there are other effects that can affect their overall resistances. We neglect those effects here for simplicity.
Example 16: The resistance of a 60.0watt lightbulb is measured 18.2Ω when cold (unlit). When connected to a120V source and 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}(^{o}C)^{1}.
Solution: P = VI or, I = P/V = 60.0w/120V = 0.500A. (when hot)
R = V/I = 120V/0.500A = 240Ω (Resistance when hot!).
Using the approximate formula R_{T} = R_{20 }[1 + a ( T  20)], we get:
240 = 18.2 [1 + 4.5x10^{3 }(T 20)] from which T = 2728 rounded to 2700 ^{o}C. The melting point of tungsten is 3400 ^{o}C.
Multiloop Circuits (Kirchhoff's Rules):
In electronics there are much more involved circuits. Circuits are often comprised of several loops. To solve for unknown currents in multiloop circuits, the socalled "Kirchhoff's rules" are used. As an example consider the 2loop circuit of Fig. 7. The loops are ABCDA and ABFEA. Verify by tracing.
Between A and B, there are 3 branches: ADCB (left), AEFB (right), and AGB (middle). Verify. The goal is to solve for the current in each branch. Name these unknown currents: I_{1 }, I_{2 }, and I_{3}_{ } . R_{1 }, R_{2 }, Battery V_{1} are in series and experience the same current I_{1}. R_{4} and Battery V_{3} also in series, experience a different current labeled I_{2}. R_{3} and Battery V_{2} experience current I_{3}. We need 3 equations to solve for the 3 unknown currents. Read the paragraphs below Fig. 7 to understand how the following 3 equations are written.

KLR in Loop AGBCDA:
(1) V_{2 } +R_{3}I_{3 }+V_{1 }+R_{2}I_{1} +R_{1}I_{1} = 0. KLR in Loop AGBFEA: (2) V_{2} + R_{3}I_{3}  R_{4}I_{2} +V_{3 } = 0. KJR at Junction A: (3)  I_{1} + I_{2} + I_{3} = 0. 
Figure 7
1) Kirchhoff's Loop Rule (KLR):
This rule states that "the sum of voltage jumps and drops across the elements in any closed loop is zero."
The voltages in a loop are either 1) due to an existing battery V_{bat} that we just write its voltage down, or 2) because of the current flowing in a resistor that we write the voltage across that resistor as the product RI. Product RI is also in Volts. In writing a KLR, we just add the battery voltages and the RI voltages for all resistors in the loop and set the total equal to zero. Of course, we need to be careful which ones are positive and which ones are negative as follows:
1) Rule for Batteries: As we trace a closed loop, clockwise for example, if we go from () to (+) across a battery, it is a voltage jump and we write +V_{bat } for that battery. If we happen to go from (+) to () of it as we trace, it is a voltage drop and we write it as V_{bat }.
2) Rule for Resistors: If our tracing direction opposes the assumed current I, we write +RI (a voltage jump) across resistor R, and if our tracing direction agrees with I, we write RI (a voltage drop) across resistor R.
The reason for the above is that when we swim against the flow of a river, we are trying to gain elevation, potential energy, or simply potential. Going from one end of a resistor to its other end against current I is like going to a higher potential or higher voltage point of the loop that means a "voltage jump." We then write the product RI with a (+) sign. If our trace direction agrees with the assumed current I, it is like losing potential and the product RI deserves a () sign.
Choosing Current Directions: We don't know the true direction of the currents to begin with. We simply assume a direction for each branch. If our assumption is correct, the answers we get at the end of our calculations will be positive. If our assumption on a current direction is wrong, calculations will yield a negative value for that current and we will know it at the end.
2) Kirchhoff's Junction Rule ( KJR ): This rule states that
"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 a point of connection of 3 or more wires. Equation (3) is a KJR written for Junction A in Fig. 7. I_{1 }leaves Junction A and is given a () sign. Both I_{2} and I_{3} go toward Junction A, they are given (+) signs.
Example 17: In the figure shown, the directions for currents I_{1}, I_{2}, and I_{3} are already chosen. Find their values and determine if the assumed directions are correct?
Figure 8
Solution: KLR in Loop ABCDA: + 40I_{3}  5 + 6 + 30I_{1} + 20 I_{1} = 0.
KLR in Loop ABFEA: + 40I_{3}  5  9 + 20I_{2} + 50 I_{2} = 0.
KJR at Junction B : + I_{1} + I_{2}  I_{3} = 0.
Rearranging the above equations to make them calculator ready, we get:
+50I_{1} + 0I_{2} + 40I_{3} = 1. 0 I_{1} +70I_{2} + 40I_{3} = 14. 1 I_{1} + 1 I_{2}  1 I_{3} = 0.

Calculation results: I_{1} =  0.0807 A. I_{2} = 0.1566 A. I_{3} = 0.0759 A. 
The direction of I_{1} was chosen incorrectly. 
The RC Circuit:
A circuit containing a resistor and a capacitor forms the "RC circuit." An RC charging circuit requires a battery as well as shown below. The resistance R is used to control the current and depending on capacity C, the charging time varies. The product RC is called the "time constant, τ"of the circuit. τ is pronounced "tau."
The unit of τ = RC is second in SI. Note: Here, I_{max} , I_{o} , and I(0) mean the same thing. 
The current in this
circuit must the same for the capacitor, resistor, and
battery, as well as the ammeter because it is a series
circuit. The
capacitor current I_{C}
decreases
exponentially with time and its equation is given by: I_{C} = I_{max } e^{(t / RC)} where I_{max} = V_{bat.}/R. The capacitor voltage V_{C } increases exponentially to its limiting value V_{bat} according to the following equation: V_{C} = V_{max}[1  e ^{(t / RC)}] where V_{max} = V_{bat.} 
Figure 9
As soon as key K is closed the battery with voltage V_{bat}_{.} 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 I_{o}. I_{o}_{ }can also be shown as I_{max}. Maximum current occurs at the very beginning at t = 0. As the capacitor fills up, it develops resistance, and therefore the battery faces an increasing resistance in excess of R. This causes the current to 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, V_{bat.} = V_{R} + V_{C }. As V_{C} increases, the current I or I_{C} goes down causing V_{R} to decrease and approach 0. The charging and discharging equations are given below:
Charging (Battery in Circuit)

At t = 0,
V_{C }= 0
I_{C }= V_{Bat}/R

As t → ∞ V_{C} → V_{Bat} _{ } I_{C }→ 0


Discharge
(Battery Removed)
Initial Charge: Q_{o}

V_{o}= Q_{o}/C

At t = 0,
V_{C }= Q_{o}/C
I_{C} = Q_{o}/RC

As t → ∞
V_{C} → 0 _{ } I_{C }→ 0

Example 18: In the figure shown, find (a) I_{o} or I_{max }, 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 values.
Solution:
Figure 10 
RC =
100Ω(0.050
Farads) = 5.00 sec. I_{max} = V_{bat.}/R = 20.0V/100Ω = 0.200A. I(t) = I_{max} e ^{(t / RC)}. I(1) = 0.2 exp (1/5) = 0.164A. I(5) = 0.2 exp (5/5) = 0.0736A. I(10) = 0.2 exp (10/5) = 0.0270A. I(100) = 0.2 exp (100/5) = 0. 
Example 19: In the previous example, calculate V_{R} and V_{C} at the given instants.
Solution:
t = 0.  V_{R} = RI = 100Ω (0.200A) = 20V.  V_{C} = V_{b} V_{R} = 20 20 = 0. 
t = 1s  V_{R} = RI = 100Ω (0.164A) = 16.4V.  V_{C} = V_{b} V_{R} = 20 16.4 = 3.6 V. 
t = 5s  V_{R} = RI = 100Ω (0.0736A) = 7.36V.  V_{C} = V_{b} V_{R} = 20 7.36 = 12.64V. 
t = 10s  V_{R} = RI = 100Ω (0.0270A) = 2.70V.  V_{C} = V_{b} V_{R} = 20 2.70 = 17.30V. 
t = 100s  V_{R} = RI = 100Ω (0 A) = 0.  V_{C} = V_{b} V_{R} = 20 0 = 20.0V. 
Example 20: In the previous example, find the values of V_{C} directly by using the equation for V_{C} that is V_{c} = V_{bat}[1  e ^{(t /}^{ RC}^{)}] and compare the results with the values of V_{C }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 ρ_{20} is the resistivity of a material at 20^{o}C, the temperature coefficient of resistivity α is (a) the change in resistivity Δρ (b) the change in resistivity per unit resistivity Δρ/ρ_{20} (c) the change in resistivity per unit resistivity per unit change in temperature Δρ/(ρ_{20}ΔT). click here.
3) If we write α as α = Δρ/(ρ_{20}ΔT) that you may want to write it with a horizontal fraction bar, it is easy to show that the unit of α is (a) ^{o}C^{1} (b) Ω/^{o}C (c) [Ω^{o}C]^{1}.
4) One of the elements used in the filament of older light bulbs was tungsten. The resistivity of tungsten (From Table 20.1) at 20^{o}C is ρ_{20} = 5.6x10^{8} Ωm and its temperature coefficient of resistivity is α = 4.5x10^{3}[^{o}C]^{1}. The resistivity of tungsten at 2720^{o}C is (a) ρ_{2720 }o_{C} = 7.3x10^{7}Ωm (b) ρ_{2720 }o_{C} = 5.9x10^{7}/^{o}C (c) neither a nor b. click here.
Problem: A student measures the resistance of an old type light bulb when cold and not in use. Its cold resistance is R_{20}o_{C} = 11Ω. If the element is made of tungsten, and we neglect the changes in its dimensions due to heat, find its resistance at 2700^{o}C. Use the approximate equation: R_{T} = R_{20}[1+ α(T20)]. Answer the following questions:
5) The resistance at 2700^{o}C of the element is (a)129Ω (b)144Ω (c)192Ω. click here.
6) Since light bulbs are designed to operate at 120V, 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 = RI^{2} (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 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^{o}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^{o}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 3V battery with its pos. pole on the left. In the cd side place an 18V battery with its neg. pole up and follow it by a 20Ω resistor. In the da side place a 15Ω 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.
16) From b to c, we to go through the 3V battery from its pos. pole to its neg. pole. This is associated with a (a) voltage drop and we write 3V (b) voltage jump and we write +3V (c) neither jump nor drop.
17) From c to d, we to go through the 18V battery from its neg. pole to its pos. pole. This is associated with a (a) voltage drop and we write 18V (b) voltage jump and we write +18V (c) neither jump nor drop. click here.
18) After the 18V battery, there comes the 20Ω resistor. Since we are following the assumed direction for the current I, we are moving toward lower potentials. The voltage change across this 20Ω resistor is therefore a drop and we write (a) 20I (b) +20I (c) 20. click here.
19) From d to a, going with the flow is associated with a (a) voltage jump across the 15Ω resistor and we write +15I (b) voltage drop across the 15Ω resistor and we write 15I (c) voltage drop across the 15Ω resistor and we write 15.
20) Going from a to b to complete the loop, we run into the 10Ω resistor only. Again, going with the assumed direction for I, there is a (a) voltage jump across the 10Ω resistor and we write +10I (b) voltage drop across the 10Ω resistor and we write 10I (c) voltage drop across the 10Ω resistor and we write 10. click here.
21) The completed KLR for Loop abcda is (a) 3+1820I15I10I (b) 3+1820I15I10I= 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 same 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. This time our travel along the loop is opposite to the assumed direction for current I. click here.
24) From b to c, we have to go through the 3V battery from its pos. pole to its neg. pole. This is associated with a (a) voltage drop and we write 3V (b) voltage jump and we write +3V (c) neither jump nor drop.
25) From c to d, we have to go through the 18V battery from its neg. pole to its pos. pole. This is associated with a (a) voltage drop and we write 18V (b) voltage jump and we write +18V (c) neither jump nor drop.
26) After the 18V battery, there comes the 20Ω resistor. Since we are going against the assumed direction for current I, we are moving toward higher potentials. The voltage change across this 20Ω resistor must be taken to be a jump and we write (a) 20I (b) +20I (c) +20. click here.
27) From d to a, going against the flow is associated with a (a) voltage jump across the 15Ω resistor and we write +15I (b) voltage drop across the 15Ω resistor and we write 15I (c) voltage jump across the 15Ω resistor and we write +15.
28) Going from a to b to complete the loop, we run into the 10Ω resistor only. Again, going against the assumed direction for I, there is a (a) voltage jump across the 10Ω resistor and we write +10I (b) voltage drop across the 10Ω resistor and we write 10I (c) voltage increase across the 10Ω 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 I_{1} and I_{3 } . Answer the following questions:
32) KLR for ABCDA is (a) 120 10 +12 90I_{1}  60I_{1} = 0 (b) 120I_{3} 10 +12 90I_{1}  60I_{1} = 0 (c) a & b.
33) KLR for ABFEA is (a) 120I_{3} 10 +18 90I_{1}  60I_{1} = 0 (b) 120I_{3} 10 18 +60I_{2} +150I_{2} = 0 (c) a & b.
34) KJR at Junction B is (a) I_{1}  I_{2} + I_{3} (b) I_{1}  I_{2} + I_{3} = 0 (c)  I_{1} + I_{2} + I_{3} = 0. click here
35) Answer for I_{1 }, I_{2 }, & I_{3} are: (a) 0.05382A, 0.10442A, 0.0506A (b) 53.82mA, 104.42mA, 50.60mA (c) a & b.
Problem: Consider the RC circuit of 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 capacitor voltage V_{C} 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 V_{C} = 0, all the battery voltage drop across (a) resistor R (b) capacitor C (c) neither a nor b.
38) The initial current I_{o} is I_{o} = V_{bat} /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 faced by the battery is R (c) neither a nor b. click here.
39) As the capacitor fills up, it develops resistance toward accepting more charges and the total resistance in the circuit keeps increasing in excess of 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.