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Questions: 1 2 3 4 5 6 7 8 9 10 11 12

Kirchhoff's Rules & RC Circuits

  1. Use the loop method to find current through each resistor in the circuit shown below.

    2.  

     
     


    Travelling around the loops in the direction shown, we find
     

    12 - 10I1 - 20(I1 + I2) = 0 (1)
    10 - 20(I2 + I1) - 5I2 = 0 (2)

    Collecting terms, these equations reduce to
     

    30I1 + 20I2 = 12 (3)
    20I1 + 25I2 = 10 (4)

    Multiplying equation (3) by 2 and equation (4) by 3 yields
     

    60I1 + 40I2 = 24 (5)
    60I1 + 75I2 = 30 (6)

    Equation (5) may be subtracted from equation (6) to yield

    (60 - 60)I1 + (75 - 40)I2 = (30 - 24) ,

    or I2 = 6/35 Amps . Substituting this result back into equation (3) yields

    30I1 + 20(6/35) = 12 ,

    or I1 = 2/7 Amps .

    Thus the current through each resistor is
     

    Resistor Current Direction
    10 Ω 6/35 A ®
    20 Ω 6/35 A + 2/7 A = 16/35 A ¯
    5 Ω 2/7 A ®

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  2. Use the loop method to find current through each resistor in the circuit shown below.

    3.  

     
     


    Travelling around the loops in the direction shown, we find
     

    10 - 20I1 - 10(I1 - I2) - 10 = 0 (1)
    10 - 10(I2 - I1) - 5I2 + 20 = 0 (2)

    Collecting terms, these equations reduce to
     

    30I1 - 10I2 = 0 (3)
    -10I1 + 15I2 = 30 (4)

    Multiplying equation (4) by 3
     

    30I1 - 10I2 = 0 (5)
    -30I1 + 45I2 = 90 (6)

    Equation (5) may be added to equation (6) to yield

    (30 + -30)I1 + (-10 + 45)I2 = 90 ,

    or I2 = 18/7 Amps . Using this result with equation (3) yields

    I1 = (1/3)I2 = 6/7 Amps .

    Thus the current through each resistor is
     

    Resistor Current Direction
    20 Ω 6/7 A ®
    10 Ω 18/7 A - 6/7 A = 12/7 A ­
    5 Ω 18/7 A ®

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  3. Use the branch method to find current through each resistor in the circuit shown below. Find the potential difference between points A and B.

    4.  

     
     


    First we assign a current to each branch as shown in the diagram above. We have two nodes, so we have only one independent current equation. From node A, we find
     

    I1 = I2 + I3 (1)

    For the two loops, following the directions shown, we get
     

    10 - 5I1 - 10I3 - 2 = 0 (2)
    15 + 10I2 - 10I3 - 2 = 0 (3)

    If we use (1) to eliminate I1 from (2), we find
     

    5I2 +15I3 = 8 (4)
    -10I2 + 10I3 = 13 (5)

    If we multiply (4) by 2 and then add the result to (5), we get
     

    40I3 = 29 (6)

    So I3 = 29/40 A. Using this result in equation (4) yields

    5I2 + 15(29/40) = 8 .

    So I2 = -23/40 A. Since I2 is negative, the true direction is reversed to that shown in the diagram. With our results for I2 and I3, (1) gives us I1 = 6/40 A.

    We can find VAB by following any of several paths
     

    VAB = 10I3 + 2 = 37/4 Volts
    = 10I2 + 15 = 37/4 Volts
    = -5I1 + 10 = 37/4 Volts

    Point A is 37/4 Volts above point B.

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  4. Use the branch method to find current through each resistor in the circuit shown below. Find the potential difference between points A and B.

    5.  

     
     


    First we assign a current to each branch as shown in the diagram above. We have four nodes, C, D, E, and F, so we have only two independent current equations. From node C and D, we find
     

    I1 = I2 + I3 (1)
    I2 = I4 + I5 (2)

    For the three loops, following the directions shown, we get
     

    1.5 - 50I1 - 15I3 = 0 (3)
    1.5 + 56I4 + 10I2 - 15I3 = 0 (4)
    -56I5 + 56I4 = 0 (5)

    Equation (5) tells us (as we might have guessed) that I4 = I5. Using this with (2) yields that I5 = ½I2. If we use (1) to eliminate I1 from (3), equations (3) and (4) reduce to
     

    1.5 - 50I2 - 65I3 = 0 (6)
    1.5 + 38I2 - 15I3 = 0 (7)

    Eliminating I3 from (6) and (7) by multiplying (6) by 3, (7) by 13 and subtracting the two yields
     

    [3(1.5) - 13(1.5)] - [3(50)+13(38)]I2 = 0 (8)

    which reduces to I2 = -15/644 A. Thus I4 = I5 = ½I2 = -15/1288 A. From equation (7), we find

    I3 = [3/2 - 38(15/644)]/15 = 33/805 A .

    And from (1), we get I1 = I2 + I3 = 57/3220 A.

    We can find VAB by following any of several paths but it is easiest to follow AGFEB

    VAB = 1.5 V - 1.5 V = 0 V .

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  5. An electronic flash attachment for a camera produces a flash by using the energy stored in a 750-μF capacitor. Between flashes, the capacitor recharges through a resistor whose resistance is chosen so that the capacitor recharges with a time constant of 3.0 s. Determine the value of the resistance.

    6.  

     
     
     
     
     

    The time constant is given by τ = RC, hence

    R = τ/C = 3.0 s / 750 μF = 4.0 kΩ .

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  6. A charged capacitor is connected across a 9600-Ω resistor and discharges to 1% of its maximum charge in a time of 8.3 s. What is the capacitance of the capacitor?

    7.  

     
     
     
     
     

    The behaviour of discharging capacitors is given by

    Q(t) = Q0e-t/RC .

    We are given that Q(t = 8.3 s) = 0.01Q0. So we have

    0.01Q0 = Q0e-t/RC.

    Eliminating Q0, and taking the natural logarithm of both sides yields

    ln(0.01) = -t/RC .

    Hence

    C = -t / Rln(0.01) = -(8.3 s)/(9600 Ω)ln(0.01) = 188 μF .

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  7. Ideal capacitors have an infinite internal resistance. Real capacitors only have a very large resistance as charges leak from one plate to the other. If a capacitor of 8.0 μF has an internal resistance of 5.0 × 108Ω, how long does it take for one-half of its original charge to leak away?

    8.  

     
     
     
     
     

    The behaviour of discharging capacitors is given by

    Q(t) = Q0e-t/RC .

    We are asked to find t such that Q(t) = ½Q0. So we have

    ½Q0 = Q0e-t/RC .

    Eliminating Q0, and taking the natural logarithm of both sides yields

    ln(0.5) = -t/RC .

    Hence

    t = -RC ln(0.5) = -(5.8 × 108Ω)(8.0 × 10-6μF)ln(0.5) = 2.77 × 103 s = 46.2 min .

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  8. Three identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 0.020 s. What is the time constant when they are connected with the same resistor as in part b?

    9.  

     
     


    The equivalent capacitance in circuit (a) is Ca = 3C/2. Thus the time constant is τa = 3RC/2 . The equivalent capacitance in circuit (b) is Cb = 2C/3. Thus the time constant is τb = 2RC/3 . Hence

    τb = (2/3)(2/3)(3RC/2) = (4/9)τa = (4/9)(0.020 s) = 0.0089 s .

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  9. In the circuit shown below,ε = 12.0 V, r = 0.500 Ω, R1 = 5.00 Ω, R2 = 10.0 Ω, and C = 250 μF. Initially, the switch S is open. (a) At the instant S is closed, determine the current supplied by the battery. (b) After the switch has be closed for a long time, determine the current supplied by the battery. (c) What is the voltage drop and charge across the capacitor at this later time? (d) The switch is now reopened, how long does it take for the capacitor to lose 80% of its charge.

    10.  

     
     


    (a) Initially, the capacitor acts like a short circuit bypassing R2. The circuit's behaviour is identical to

    Thus the current through the series combination is

    I = ε/(r + R1) = (12.0 V) / (5.5 Ω) = 2.18 A .

    (b) After a long time the capacitors acts as an open switch and all the resistors are in series. The circuit is now identical to

    Thus the current through the series combination is

    I = ε/(r + R1 + R2) = (12.0 V)/(15.5 Ω) = 0.774 A .

    (c) The capacitor is in parallel with R2, so the must both have the same voltage drop. From Ohm's Law, we find the voltage drop to be

    VC = V2 = IR2 = (0.774 A)(10.0 Ω) = 7.74 V .

    The charge on the capacitor is then given by

    Q = CVc = (250 μF)(7.74 V) = 1.94 mC .

    (d) The behaviour of discharging capacitors is given by

    Q(t) = Q0e-t/RC .

    We are asked to find t such that Q(t) = 0.2Q0. So we have

    0.2Q0 = Q0e-t/RC .

    Eliminating Q0, and taking the natural logarithm of both sides yields

    ln(0.2) = -t/RC .

    Hence

    t = -RC ln(0.2) = -(10.0 Ω) (250 μF) ln(0.2) = 4.02 × 10-3 s .

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  10. A galvanometer has a coil resistance of 250 Ω and requires a current of 1.5 mA for full-scale deflection. This device is used in an ammeter that has a full-scale deflection of 25.0 mA. What is the value of the shunt resistance?

    11.  

     
     
     
     
     

    The shunt resistance is the resistor connected in parallel with the coil as shown in the diagram below. The current entering and leaving the branch is I = 25.0 mA. We want the current through the coil to be IG = 1.5 mA.

    Since the voltage drop across each arm is the same

    IGRC = (I - IG)RS .

    Solving for RS, we find

    RS = IGRC / (I - IG) = (1.5 mA)(250 Ω)/(25 mA - 1.5 mA) = 16 .

    We would need the shunt to have a resistance of 16 Ω.

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  11. The coil resistor in an ammeter has a resistance which is 100 times larger than the shunt resistor. The galvanometer reads 10.0 mA when the ammeter is used to measure the current in a simple circuit. Unfortunately, the resistor in the simple circuit has a resistance which is only 5.00 times as large as the shunt resistor. What would be current through the resistor if the ammeter was not in place?

    12.  

     
     


    With the ammeter in place, the current produced by the battery is I = V/Req, where

    Req = 5R + (1/R + 1/100R)-1 = (605/101)R .

    Thus
     

    I = (101/605)(V/R) (1)

    When the ammeter is not in the circuit the current will be
     

    I0 = V/5R (2)

    We can use equation (1) to eliminate V/R from equation (2),
     

    I0 = (1/5)(605/101)I = (121/105)I (3)

    So to find I0 we need to know I. Looking at the parallel arms of the ammeter we see that the voltage drop must be the same over each arm

    IG(100R) = (I - IG)R .

    Solving for I yields I = 101IG. Using this result with (3) gives

    I0 = (121/105)I = (121/105)(101)(10 × 10-3 A) = 1.22 A .

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  12. A galvanometer with a full-scale deflection of 2000 μA has a coil resistance of 100 Ω. If it is to be used as a voltmeter with a full-scale deflection of 1.5 V, what would be the required multiplier resistance?

    13.  

     
     
     
     
     

    A voltmeter is connected is parallel with the resistor of interest, as shown in the diagram below

    Since the voltmeter is parallel with the resistor, the voltage drop across both is the same. The voltage drop across the coil is

    Vcoil = (RM + RC)IG .

    Solving for RM, yields

    RM = V/IG - RG = (1.5 V / 2000 A) - 100 Ω = 650 Ω .

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