PHYS 1220 Kirchhoff's Rules & RC Circuits

Kirchhoff's Rules & RC Circuits

Give the minimum number of equations to find the current in each branch of the circuit shown below. Find the currents.

There are three branches, so we will have three unknown currents as shown above. There are two loops with no interior branches, so we will get two voltage equations. The final equation comes from writing a current equation at one of the nodes, point A or B above.

12 – 10I₁ – 20I₃	=	0	(1)
10 – 20I₃ – 5I₂	=	0	(2)
I₁ + I₂	=	I₃	(3)

Bringing the constants to one side, we have a system of equations to solve.

I₁	I₂	I₃
10	0	20	12
0	5	20	10
1	1	–1	0

From a spreadsheet, or any other method, we find

*Resistor*	*Current*	*Direction*
10 Ω	I₁ = 2/7 A	→
20 Ω	I₃ = 16/35 A	↓
5 Ω	I₂ = 6/35 A	→

The power supplied by the batteries is P_in = 12 V × 2/7 A + 10 V × 6/35 A = 36/7 W.

The Joule heating of the resistors is

P_out = (2/7 A)²(10 Ω) + (16/35 A)²(20 Ω) + (6/35 A)²(5 Ω) = 36/7 W.

So P_in = P_out confirming the correctness of our solution.

Give the minimum number of equations to find the current in each branch of the circuit shown below. Find the currents.

10 – 20I₁ – 10I₃ – 10	=	0	(1)
– 5I₂ + 20 + 10 + 10I₃	=	0	(2)
I₁	=	I₂ + I₃	(3)

Bringing the constants to one side, we have a system of equations to solve.

I₁

I₂

I₃

20

0

10

0

0

5

–10

30

1

–1

–1

0

From a spreadsheet, or any other method, we find

*Resistor*	*Current*	*Direction*
20 Ω	I₁ = 6/7 A	→
10 Ω	I₃ = –12/7 A	↑ (guessed wrong)
5 Ω	I₂ = 18/7 A	→

Since I₃ is negative, the true direction is reversed to that shown in the diagram.

The power supplied by the batteries is

P_in = 10 V × 6/7 A + 10 V × 12/7 A + 20 V × 18/7 A = 540/7 W .

The Joule heating of the resistors is

P_out = (6/7 A)²(20 Ω) + (12/7 A)²(10 Ω) + (18/7 A)²(5 Ω) = 540/7 W.

So P_in = P_out confirming the correctness of our solution.

Give the minimum number of equations to find the current in each branch of the circuit shown below. Find the currents. Find the potential difference between points A and B.

10 – 5I₁ – 10I₃ – 2	= 0	(1)
15 + 10I₂ – 10I₃ – 2	= 0	(2)
I₁	= I₂ + I₃	(3)

Bringing the constants to one side, we have a system of equations to solve.

I₁

I₂

I₃

5

0

10

8

0

–10

10

13

1

–1

–1

0

From a spreadsheet, or any other method, we find

Resistor

Current

Direction

5 Ω

I₁ = 6/40 A

↑

10 Ω

I₃ = 29/40 A
↑

10 Ω;

I₂ = –23/40 A

← (guessed wrong)

Since I₂ is negative, the true direction is reversed to that shown in the diagram.

The power supplied by the batteries is

P_in = 10 V × 6/40 A + 15 V ×23/40 A − 2 V × 29/40 A = 347/40 W .

Note that the 10 V and 15 V batteries supply energy to the circuit while the 2 V battery is charging or taking energy out of the circuit (hence the minus sign above).

The Joule heating of the resistors is

P_out = (6/40 A)²(5 Ω) + (23/40 A)²(10 Ω) + 29/40 A)²(10 Ω) = 347/40 W.

So P_in = P_out confirming the correctness of our solution.

We can find V_AB by following any of several paths

V_AB	= 10I₃ + 2	= 37/4 Volts
	= 10I₂ + 15	= 37/4 Volts
	= –5I₁ + 10	= 37/4 Volts

Point A is 37/4 Volts above point B.

Give the minimum number of equations to find the current in each branch of the circuit shown below. Find the currents. Find the potential difference between points A and B.

We have four nodes, C, D, E, and F above. We assign a current of arbitrary direction to each branch as shown in the diagram above. There are six branches and therefore six unknown currents, so we need six equations. There are three loops with no interior branches which will yield three voltage equations. For the other three equations, we apply the current law to three of the nodes as shown below.

(C)	I₁	=	I₂ + I₃	(1)
(D)	I₂	=	I₄ + I₅	(2)
(E)	I₄ + I₅	=	I₆	(3)

For the three loops, we get

1.5 – 50I₁ – 15I₃	= 0	(4)
1.5 + 56I₄ + 10I₂ – 15I₃	= 0	(5)
–56I₅ + 56I₄	= 0	(6)

Bringing the constants to one side, we have a system of six equations to solve.

I₁	I₂	I₃	I₄	I₅	I₆
1	–1	–1	0	0	0	0
0	1	0	–1	–1	0	0
0	0	0	1	1	–1	0
50	0	15	0	0	0	1.5
0	10	–15	56	0	0	–1.5
0	0	0	56	–56	0	0

We find

*Resistor*	*Current*	*Direction*
50 Ω	I₁ = 57/3220 A	→
10 Ω	I₂ = –15/644 A	← (guessed wrong)
15 Ω	I₃ = 33/805 A	↓
56 Ω	I₄ = –15/1288 A	↑ (guessed wrong)
56 Ω	I₅ = –15/1288 A	↑ (guessed wrong)
none	I₆ = –15/644 A	→ (guessed wrong)

The power supplied by the batteries is

P_in = 1.5 V × 57/3220 A + 1.5 V × 15/644 A = 198/3220 W.

Note that the direction of I₆ is actually in the direction opposite to shown and thus the bottom battery supplies energy to the circuit.

The Joule heating of the resistors is

P_out = (57/3220)²(50) + (15/644)²(10) + (33/805)²(15) + (15/1288)²(56 + 56) = 198/3220 W.

So P_in = P_out confirming the correctness of our solution.

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

V_AB = 1.5 V − 1.5 V = 0 V.

For the circuit shown below give the minimum set of equations that determines the current in each branch. You do not need to solve the system of equations.

We have four nodes, A, B, C, and D above. We assign a current of arbitrary direction to each branch as shown in the diagram above. There are six branches and therefore six unknown currents, so we need six equations. There are three loops with no interior branches which will yield three voltage equations. For the other three equations, we apply the current law to three of the nodes as shown below.

(A)	I₁+ I₂ =	I₄	(1)
(B)	I₃ + I₅ =	I₁	(2)
(C)	I₄=	I₅ + I₆	(3)

For the three loops, we get

10 – 25I₄ – 50I₅	= 0	(4)
6 – 60I₂ – 25I₄ – 100I₆	= 0	(5)
4 – 30I₃ + 3 + 50I₅ – 100I₆	= 0	(6)

Bringing the constants to one side, we have a system of six equations to solve.

I₁	I₂	I₃	I₄	I₅	I₆
1	1	0	–1	0	0	0
–1	0	1	0	1	0	0
0	0	0	1	–1	–1	0
0	0	0	25	50	0	10
0	60	0	25	0	100	6
0	0	30	0	–50	100	7

We find

*Resistor*	*Current*	*Direction*
none	I₁ = 1033/3690 A	As shown
60 Ω	I₂ = –361/3690 A	→ (guessed wrong)
30 Ω	I₃ = 631/3690 A	←
25 Ω	I₄ = 672/3690 A	↓
50 Ω	I₅ = 402/3690 A	←
100 Ω	I₆ = 270/3690 A	→

The power supplied by the batteries is

P_in = 10 V × 1033/3690 A – 6 V × 361/3690 A + (3 V + 4 V) × 631/3690 A = 12581/3690 W.

Note that the direction of I₂ is actually in the direction opposite to shown and thus the 6 V battery is being charged.

The Joule heating of the resistors is

P_out = (361/3690)²(60) + (631/3690)²(30) + (672/3690)²(25) + (402/3690)²(50) + (270/3690)²(100)

= 12581/3690 W.

So P_in = P_out confirming the correctness of our solution.

Consider the circuit below. Determine the potential with respect to the negative terminal of the battery at the points A, B, and C labelled on the diagram.

Since the resistors are all in series we can use the voltage divider equation

Consider the circuit below. The potential with respect to the negative terminal of the battery at point B is 6.0 V. Determine the battery voltage ε and the potential at the points A and C labelled on the diagram.

Since the resistors are all in series we can use the voltage divider equation

Solving for the battery voltage ε, we find ε = 8.5714 Volts. Hence we find V_A and V_C

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?

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 I_G = 1.5 mA.

Since the voltage drop across each arm is the same

I_GR_C = (I − I_G)R_S .

Solving for R_S, we find

R_S = I_GR_C / (I − I_G) = (1.5 mA)(250 Ω)/(25 mA − 1.5 mA) = 16 .

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

Consider the circuit in diagram (a) below. What is the current in this circuit? Now consider the circuit in diagram (b) below. The only difference is that an ammeter has been added so that the current could be measured. The ammeter is the same one as the previous problem. What is the current reading on the ammeter? Why is it different from the theoretical value that you found for diagram (a)?

For diagram (a) we just use Ohm's law to determine the current

I = V/R = (0.6 V)/(20 W) = 0.030 A = 30 mA.

For diagram (b) we must remember that this ammeter has a resistance R_A = 15 W which is not much smaller that the resistance already there 20 Ω. The equivalent resistance of the circuit in (b) is therefore

R_eq = 15 Ω + 20 Ω = 35 Ω.

Now using Ohm's law to determine the current

I_A = V/R_eq = (0.6 V)/(35 Ω) = 0.017 A = 17 mA .

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?

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

R_eq = 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

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

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

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

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?

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

V_coil = (R_M + R_C)I_G .

Solving for R_M, yields

R_M = V/I_G - R_G = (1.5 V / 2000 A) - 100 Ω = 650 Ω .

Consider the circuit in diagram (a) below. What is the potential difference over the 140 kΩ resistor in this circuit? Now consider the circuit in diagram (b) below. The only difference is that a voltmeter has been added so that the potential difference could be measured. The voltmeter is the same one as the previous problem. What is the voltage reading on the voltmeter? Why is it different from the theoretical value that you found for diagram (a)?

For diagram (a) we can determine the potential difference over the 140 kΩ resistor quite simply. It is twice as big as the 70 kW resistor. Hence it uses twice as much energy and so has twice the potential difference. That is it gets 2/3 × 9 Volts = 6 Volts.

For diagram (b) we must remember that this voltmeter has a resistance R_V = 140 k&Omeg a which is not much greater that the resistance it is connected in parallel with, 140 k Ω. The equivalent resistance of the voltmeter and the 140 kΩ resistor in (b) is therefore

R_eq = (1/(140 kΩ) + 1/(140 kΩ)^-1 = 70 k Ω.

This equivalent resistance is the same size as the other resistor in circuit so each shares the half the potential difference or ½ × 9 Volts = 4.5 Volts. Now recall that the equivalent resistance and the two parallel resistors must each have the same potential difference. So the 140 kΩ resistor has 4.5 Volts over it and that is what the voltmeter displays.

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.

The time constant is given by τ = RC, hence

R = τ/C = 3.0 s / 750 μF = 4.0 kΩ .

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?

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 .

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?

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 .

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?

The equivalent capacitance in circuit (a) is C_a = 3C/2. Thus the time constant is τ_a = 3RC/2 . The equivalent capacitance in circuit (b) is C_b = 2C/3. Thus the time constant is τ_b = 2RC/3 . Hence

τ_b = (2/3)(2/3)(3RC/2) = (4/9)τ_a = (4/9)(0.020 s) = 0.0089 s .

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.

(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

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

The charge on the capacitor is then given by

Q = CV_c = (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 .

Consider the circuit below. Both capacitors are initially uncharged. Switch S₂ is closed followed by S₁.

(a)At that instant what is the conventional current (magnitude and direction) in each resistor?

(b) >After the switches have been closed for a long time, what is the conventional current (magnitude and direction) in each resistor?

(c) Now S₂ (and only S₂) is reopened. At this instant what is the conventional current (magnitude and direction) in each resistor?

(d)How long will it take for the current in the 30 Ω resistor to drop to 0.10 A?

(a) At t = 0, capacitors act as shorts (a straight wire with no resistance). Thus each if the resistors is in parallel with the 12 V battery. The conventional current will be up in each case.

I₁₅ = 12 V / 15 ΩΩ = 0.8 A, I₂₀ = 12 V / 20 Ω = 0.6 A, and I₃₀ = 12 V / 30 Ω = 0.4 A

(b) At t = , the capacitors act as opens and no current will flow through the branches containing them, that is I₁₅ = I₃₀ = 0 A. The 20 Ω resistor will still be in parallel with the battery and will still carry I₂₀ = 12 V / 20 Ω = 0.6 A upwards for conventional current.

(c) When S₂ is reopened, the 50 μF capacitor will discharge through the 20 Ω and 30 Ω resistors which are in series. To determine the current, note that in the previous question that the capacitor was fully charged and that it and the 30 Ω resistor were in parallel with the 12 V battery. Since there was no current through the resistor, the capacitor was charged to 12 V. That means now the 12 V across the capacitor will discharge through the two resistors that are in series and act as a single 50 Ω resistor.

I₂₀ = I₃₀ = 12 V / 50 Ω = 0.24 A.

The conventional current will clockwise through the loop.

(d) We know I = I₀e^-t/RC, for a discharging capacitor, where R is the equivalent resistance of the circuit which is 50 Ω and I₀ = 0.24 A from the previous question. Inverting the equation to find t yields,

t = RC ln(I₀/I) = 50 Ω × 50 μF × ln(0.24 / 0.10) = 2.2 × 10^-3 seconds.

When the switch is closed in the circuit below what is the initial current in each resistor? After the switch has been closed for a long time what is the current through each resistor? What is the voltage across the capacitor and its charge at that later time? If the switch S is reopened, what is the current through each resistor? How long will it take for the capacitor to lose 70% of its charge (Hint: what is the equivalent resistance of circuit)?

(a) At t = 0, capacitors act as shorts (a straight wire with no resistance). Thus the 1.0 Ω and 10.0 Ω resistors are in parallel and similarly the 9.0 Ω and 5.0 Ω are in parallel. These pairs can be replaced by their equivalent resistors, 0.90909 Ω and 3.21429 Ω respectively, as shown below.

⇒

Using the voltage divider formula, the voltage across the 3.21429 Ω resistor is

12.0 V × 3.21429 Ω / (0.90909 Ω + 3.21429 Ω) = 9.35434 Volts.

This is also the voltage across both the 5.0 Ω and 9.0 Ω resistors, so using Ohm’s Law

I₉ = 9.35434 V / 9 Ω = 1.03937 A and I₅ = 9.35434 V / 5 Ω = 1.87087 A

The voltage drop across the 0.90909 Ω resistor is 12.0 V – 9.35434 V = 2.64566 V. This is also the voltage across the 1.0 Ω and 10.0 Ω resistors. Again using Ohm’s Law,

I₁ = 2.64566 V / 1 Ω = 2.64566 A and I₁₀ = 2.64566 V / 10 Ω = 0.26457 A.

(b) At t = ∞, the capacitor acts as an open and no current will flow through that branch. The 1.0 Ω and 9.0 Ω resistor are now in series in one branch with the 10.0 Ω and 5.0 Ω in series in the other as shown below.

We can use the voltage divider equation to get the drop over the 9.0 Ω resistor,

12.0 V × 9.0 Ω / (1.0 Ω + 9.0 Ω) = 10.8 V.

Which means the voltage drop over the 1.0 Ω resistor is 12.0 V – 10.8 V = 1.2 V.

Similarly the voltage drop over the 5.0 Ω resistor is

12.0 V × 5.0 Ω / (10.0 Ω + 5.0 Ω) = 4.0 V

and the remaining voltage drop over the 10.0 Ω resistor is 12.0 V – 4.0 V = 8.0 V.

Using Ohm’s Law the current through each resistor is

I₁ = 1.2 V / 1 Ω = 1.2 A, I₉ = 10.8 V / 9 Ω = 1.2 A,

I₁₀ = 8.0 V / 10 Ω = 0.8 A, and I₅ = 4.0 V / 5 Ω = 0.8 A.

(c) To find the voltage drop across the capacitor, we apply Kirchhoff’s voltage rule to any loop that contains the capacitor. Going clockwise from the capacitor

V_C + (1.0 Ω) I₁ – (10.0 Ω)I₁₀ = 0.

Using the currents from the previous question, V_C = 8 V – 1.2 V = 6.8 V.

(d) When switch S is reopened the active circuit is

Kirchhoff's voltage rule for the top loop is 6.8 V – I_top(1.0 Ω + 10.0 Ω) = 0, which means the current through these resistors is 0.61818 A. Similarly, Kirchhoff’s voltage rule for the bottom loop is 6.8 V – I_bottom(9.0 Ω + 5.0 Ω) = 0, which means the current through these resistors is 0.48571 A.

(e) First, as suggested, let’s find the equivalent resistance. The top branch has a series equivalent resistance of 11.0 Ω while the bottom branch has a series equivalent resistance of 14.0 Ω. The top and bottom branch are in parallel, so they may be replaced by a single equivalent resistor

R = (1 / 11 Ω + 1/ 14 Ω)^-1 = 6.16 Ω.

We know I = I₀e^-t/RC for a discharging capacitor, where R is the equivalent resistance of the circuit which is 6.16 Ω. Inverting the equation to find t yields,

t = RC ln(I₀/I) = 6.16 Ω × 2.2 μF × ln(1 / 0.70) = 4.83 × 10^-6 seconds.

Consider the RC circuits shown below. The voltage drop and its direction for the capacitor are given. Find the current and its direction at the instant the switch S is closed.

We apply Kirchhoff’s voltage rule to each circuit being careful to note the polarity of the battery and the capacitor. Let’s assume the current is counterclockwise in each circuit and go around the loop in the same direction.

(a) 5 V + 3 V – I(10) = 0. Thus I = 0.8 A counterclockwise.

(b) 5 V – 3 V – I(10) = 0. Thus I = 0.2 A counterclockwise.

(d) 5 V – 10 V – I(10) = 0. Thus I = –0.5 A (– means clockwise).

(e) 5 V + 5 V – I(10) = 0. Thus I = 1.0 A counterclockwise.

(f) 5 V – 5 V – I(10) = 0. Thus I = 0 A.

Questions?mike.coombes@kwantlen.ca