Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

PHYS 2420 DC Circuits Solutions

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



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)

The above set of equations may be solved using MAPLE to yield I₁ = 2/7 Amps and I₂ = 6/35 Amps. Note that we are not finished as these are not the currents through the resistors.

The current through each resistor is

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

It is always wise to check the results by making sure that the power supplied by the batteries matches the Joule heating of the resistors.

Each battery supplies energy to the circuit so the power in is

P_in = ε₁I₁ + ε₂I₂ = (12 V)(2/7 A) + (10 V)(6/35 A) = 180/35 W.

As expected, the resistors consume

Pout	= IA2RA + IB2RB + IC2RC
	= (2/7 A)2(10 Ω) + (16/35 A)2(20 Ω) + (6/35 A)2(5 Ω)
	= 180/35 W

So our results are consistent.



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



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)

The above set of equations may be solved using MAPLE to yield I₁ = 6/7 Amps and I₂ = 18/7 Amps. Note that we are not finished, as these are not the currents through the resistors.

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	®

Next we check the results by making sure that the power supplied by the batteries matches the Joule heating of the resistors.

Each battery supplies energy to the circuit, so the power in is

Pin	= ε1I1 + ε2I2 + ε3(I1+I2)
	= (10 V)(6/7 A) + (20 V)(18/7 A) + (10 V)(12/7 A)
	= 540/7 W.

As expected, the resistors consume

Pout	= IA2RA + IB2RB + IC2RC
	= (6/7 A)2(20 Ω) + (12/7 A)2(10 Ω) + (18/7 A)2(5 Ω)
	= 540/7 W

So our results are consistent.



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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.



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)

These last two equations can be rearranged to yield

5I2 +10I3	= 8	(2a)
-10I2 + 10I3	= 13	(3a)

The equations may be solved using MAPLE to yield I₁ = 6/40 Amps, I₂ = -23/40 Amps, and I₃ = 29/40 A. Since I₂ is negative, the true current direction is reversed to that shown in the diagram.

As usual, we should check our results. The current directions indicate that the 10 V and 15 V battery each supply energy to the circuit while the 2 V battery is taking energy (being charged), so the power in is

P_in = ε₁I₁ + ε₂I₂ - ε₃I₃ = (10 V)(6/40 A) + (15 V)(23/40 A) - (2 V)(29/40 A) = 347/40 W.

As expected, the resistors consume

Pout	= I12RA + I22RB + I32RC
	= (6/40 A)2(5 Ω) + (23/40 A)2(10 Ω) + (29/40 A)2(10 Ω)
	= 347/40 W

We can find V_AB 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.



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 three independent current equations. From node C, D, and E, we find

I1	= I2 + I3	(1)
I2	= I4 + I5	(2)
I6 + I4 + I5	= 0	(3)

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

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

The equations may be solved using MAPLE to yield I₁ = 57/3220 A, I₂ = -15/644 A, I₃ = 33/805 A, I₄ = I₅ = -15/1288 A, and I₆ = 15/644 A. Since I₂, I₄, and I₅ are negative, the true current directions are reversed to that shown in the diagram.

As usual, we should check our results. The current directions indicate that the batteries supply energy to the circuit. The power in is

P_in = ε₁I₁ + ε₂I₆ = (1.5 V)(57/3220 A) + (1.5 V)(15/644 A) = 198/3220 W.

As expected, the resistors consume

Pout	= I12RA + I22RB + I32RC + I42RD + I52RE
	= (57/3220 A)2(50 Ω) + (15/644 A)2(10 Ω)    + (33/805 A)2(15 Ω) + (15/1288 A)2(56 Ω)    + (15/1288 A)2(56 Ω)
	= 198/3220 W

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.



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5. Redo question #1 using the superposition principle.

The superposition principle says that the current through a resistor because of one battery is unaffected by the presence of any other battery. That means if we find the current through each resistor in the following two circuits then the sum of the currents should give the same result as question #1.



These circuits are relatively easy to solve since we have only one battery and resistors in series and parallel.

The equivalent resistance of the circuit on the left, the one with the 12 V battery, is R_A = (1/5 Ω + 1/20 Ω)^-1 + 10 Ω = 4 Ω +10 Ω = 14 Ω. Using Ohm's Law, the current from the battery is I_A = V_A/R_A = 12 V/ 14 Ω = 6/7 A. This is also the current through the 10 Ωresistor, I_10A. This current splits when it reaches the parallel resistors. The current through the 20 Ωresistor will be I_20A = (4/20)I_10A = 6/35 A. The current through the 5 Ωresistor will be I_5A = (4/5)I_10A = 24/35 A.

The equivalent resistance of the circuit on the right, the one with the 10 V battery, is R_B = (1/10 Ω + 1/20 Ω)^-1 + 5 Ω = 20/3Ω + 5 Ω = 35/3 Ω. Using Ohm's Law, the current from the battery is I_B = V_B/R_B = 10 V / (35/3 Ω) = 6/7 A. This is also the current through the 5 Ωresistor, I_5B. This current splits when it reaches the parallel resistors. The current through the 20 Ω resistor will be I_20B = (20/3 / 20)I_5B = 2/7 A. The current through the 10 Ω resistor will be I_10B = (20/3 /10)I_5B = 4/7 A.

Thus the total current through each resistor, paying careful attention to current direction, is

Resistor	Current	Direction
10 Ω	6/7 A - 4/7 A = 2/7 A	®
20 Ω	6/35 A + 2/7 A = 16/35 A	¯
5 Ω	6/7 A - 24/35 A = 6/35 A	®

As expected, these are the same results as question 1.



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6. Redo question # 2 using the superposition principle.

The superposition principle says that the current through a resistor because of one battery is unaffected by the presence of any other battery. That means if we find the current through each resistor in the following three circuits then the sum of the currents should give the same result as question #2.



These circuits are relatively easy to solve since we have only one battery and resistors in series and parallel.

The equivalent resistance of the circuit on the left, the one with the 10 V battery, is R_A = (1/5 Ω + 1/10 Ω)^-1 + 20 Ω = 10/3 Ω +10 Ω = 70/3 Ω. Using Ohm's Law, the current from the battery is I_A = V_A/R_A = 10 V/ (70/3 Ω) = 3/7 A. This is also the current through the 20 Ωresistor, I_20A. This current splits when it reaches the parallel resistors. The current through the 10 Ωresistor will be I_10A = (10/3 / 10)I_20A = 1/7 A. The current through the 5 Ωresistor will be I_5A = (10/3 / 5)I_20A = 2/7 A.

The equivalent resistance of the circuit in the middle, the one with the 10 V battery, is R_B = (1/5 Ω + 1/20 Ω)^-1 + 10 Ω = 4 Ω +10 Ω = 14 Ω. Note that the 5 Ωand 20 Ωresistors share two common points and thus are in parallel. Using Ohm's Law, the current from the battery is I_B = V_B/R_B = 10 V/ 14 Ω = 5/7 A. This is also the current through the 10 Ωresistor, I_10B. This current splits when it reaches the parallel resistors. The current through the 20 Ωresistor will be I_20B = (4/20)I_10B = 1/7 A. The current through the 5 Ωresistor will be I_5B = (4/5)I_10B = 4/7 A.

The equivalent resistance of the circuit on the right, the one with the 20 V battery, is R_C = (1/10 Ω + 1/20 Ω)^-1 + 5 Ω = 20/3 Ω + 5 Ω = 35/3 Ω. Using Ohm's Law, the current from the battery is I_C = V_C/R_C = 20 V / (35/3 Ω) = 12/7 A. This is also the current through the 5 Ωresistor, I_5C. This current splits when it reaches the parallel resistors. The current through the 20 Ωresistor will be I_20C = (20/3 / 20)I_5C = 4/7 A. The current through the 10 Ωresistor will be I_10C = (20/3 /10)I_5C = 8/7 A.

Thus the total current through each resistor, paying careful attention to current direction, is

Resistor	Current	Direction
20 Ω	3/7 A - 1/7 A + 4/7 A = 6/7 A	®
10 Ω	-1/7 A + 5/7 A + 8/7 A = 12/7 A	­
5 Ω	2/7 A + 4/7 A + 12/7 A = 18/7 A	®

As expected, these are the same results as question 2.



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7. 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 .



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8. In the figure below, C₁ = C₅ = 3.00 μF and C₂ = C₃ = C₄ = 2.00 μF. What is the equivalent capacitance of the circuit? A potential of 600 V is applied across points A and B. What is the charge on each capacitor? What is the energy stored in each capacitor?



Capacitors C₃ and C₄ are in series and therefore 1/C₃₄ = 1/2 + 1/2 = 1, or C₃₄ = 1 μF.



Capacitors C₂ and C₃₄ are in parallel and therefore C₂₃₄ = 2 + 1 = 3 μF.



Final C₁ and C₂₃₄ and C₅ are is series and therefore 1/C₁₂₃₄₅ = 1/3 + 1/3 +1/3 = 1, or C₁₂₃₄₅ = 1 μF. The equivalent capacitance is 1 μF.



For each capacitor Q = CV, and we need to recall the following:

Series	Parallel
1/Cs = 1/C1 + 1/C2	Cp = C1 + C2
Same charge on Cs, C1, and C2	Same Voltage
Different voltage	Different charge on Cp, C1, and C2

Then we work backwards from the equivalent capacitor. Since V = 600 Volts and C₁₂₃₄₅ = 1 μF, the charge on the equivalent capacitor is Q₁₂₃₄₅ = 600 μC. The energy is U₁₂₃₄₅ = 1/2QV = 0.180 J.



The capacitor C₁₂₃₄₅ is actually C₁, C₂₃₄, and C₅ in series, so each has the same charge of 600 μC. The voltage drop over each is V₁ = V₂₃₄ = V₅ = Q/C = 600 μC / 3 μF = 200 V. The energy for each using U = 1/2QV yields U₁ = U₂₃₄ = U₅ = 0.060 J.



The capacitor C₂₃₄ is actually C₂ and C₃₄ in parallel, so there is 200 volts over each. The charge on C₂ is Q₂ = C₂V = 400 μC. The charge on C₃₄ is Q₃₄ = C₃₄V = 200 μC. The energy for each using U = 1/2QV yields U₂ = 0.040 J and U₃₄ = 0.020 J.



The capacitor C₃₄ is actually C₃ and C₄ in series, so each has the same charge of 200 μC. The voltage drop over each is V₂ = V₄ = Q/C = 200 μF / 2 μF = 100 V. The energy for each using U = 1/2QV yields U₂ = U₄ = 0.010 J.



 



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9. What is the equivalent capacitance of the circuit shown below? A potential of 24 V is applied across points A and B. What is the charge on each capacitor? What is the energy stored in each capacitor?



The 1 μF and 3 μF capacitors are in parallel and are equivalent to a single 4 μF capacitor. The 6 μF and 2 μF capacitors are also in parallel and are equivalent to a single 8 μF capacitor.



The two 4 μF capacitors are in series, therefore 1/C_p = 1/4 + 1/4 = 1/2, or C_p = 2 μF. The two 8 μF capacitors are in series, therefore 1/C_p = 1/8 + 1/8 = 1/4, or C_p = 4 μF.



The 2 μF and 4 μF capacitors are in parallel and are equivalent to a single 6 μF capacitor. The equivalent capacitance is 6 μF.



For each capacitor Q = CV, and we need to recall the following:

Series	Parallel
1/Cs = 1/C1 + 1/C2	Cp = C1 + C2
Same charge on Cs, C1, and C2	Same Voltage
Different voltage	Different charge on Cp, C1, and C2

Then we work backwards from the equivalent capacitor. Since V = 24 Volts and C_eq = 6 μF, the charge on the equivalent capacitor is Q_eq = 24 V ´ 6 μF = 144 μC. The energy is U_eq = 1/2QV = 1.728 mJ.

 



The equivalent capacitor is actually a 2 μF and a 4 μF capacitor in parallel, so each has the same voltage of 24 V. The charge on the 2 μF capacitor is Q = 2μF ´ 24 V = 48 μC. The charge on the 4 μF capacitor is Q = 4μF ´ 24 V = 96 μC. The energy for each using U = 1/2QV yields U₂ = 0.576 mJ and U₄ = 1.152 mJ.



The 2 μF capacitor is actually two 4 μF capacitors in series, so each has the same charge of 48 μC. The voltage drop over each is V = Q/C = 48 μC / 4 μF = 12 V. The energy for each using U = 1/2QV yields U_4R = U_4L = 0.288 mJ.

The 4 μF capacitor is actually two 8 μF capacitors in series, so each has the same charge of 96 μC. The voltage drop over each is V = Q/C = 96 μC / 8 μF = 12 V. The energy for each using U = 1/2QV yields U_8R = U_8L = 0.576 mJ.



The top right 4μF capacitor is actually a 1 μF and a 3 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 1 μF capacitor is Q = 1μF ´ 12 V = 12 μC. The charge on the 3 μF capacitor is Q = 3μF ´ 12 V = 36 μC. The energy for each using U = 1/2QV yields U₁ = 0.072 mJ and U₃ = 0.216 mJ.

The bottom left 8μF capacitor is actually a 6 μF and a 2 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 6 μF capacitor is Q = 6μF ´ 12 V = 72 μC. The charge on the 2 μF capacitor is Q = 2μF ´ 12 V = 24 μC. The energy for each using U = 1/2QV yields U₆ = 0.432 mJ and U₂ = 0.144 mJ.

 





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10. 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.



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11. A charged capacitor is connected across a 9600-Ωresistor 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 / R ln(0.01) = -(8.3 s)/(9600 Ω)ln(0.01) = 188 μF .



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12. 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 ´ 10⁸ Ω, 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) = 1/2Q₀. So we have

1/2Q₀ = 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 ´ 10⁸ Ω)(8.0 ´ 10^-6 F) ln(0.5) = 2.77 ´ 10³ s = 46.2 min .



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13. 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

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



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14. 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 and there is no charge on the capacitor. (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) Obtain expressions for the current through each resistor and for the charge on the capacitor as a function of time, that is use Kirchhoff's rules and use MAPLE to solve the system of differential equations. Remember that i = dq/dt. (e) The switch is now reopened, how long does it take for the capacitor to lose 80% of its charge.



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 .

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 .

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 .

First we assign a current to each branch and then apply Kirchhoff's Rules.



Our equations are

i1	= i2 + i3	(1)
e - i1(R1+r) - i2R2	= 0	(2)
e - i1(R1+r) - q/C	=0	(3)

Note that i₃ = dq/dt. Our initial conditions are q(0) = 0, since the capacitor is uncharged before the switch is thrown. Making use of MAPLE, the solutions are:



In the limit t ® 0, these become q(0) = 0, i₁(0) = ε/(R₁ + r), i₂(0) = 0 as we have seen in part (a). In the limit t ® ¥ , these become q(¥ ) = 0 and i₁(¥ ) = i₂(¥ ) = ε/(R₂ + R₁ + r) as we have seen in part (b).

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) = -(15.5 Ω)(250 μF) ln(0.2) = 4.02 ´ 10^-3 s .



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15. The inductors in the circuit shown below are magnetically shielded from one another so that they do not produce flux in one another. Determine the time constant of the circuit. Determine how long it takes the current through the resistor to reach 85% of its maximum. At this point, what is the energy stored in each inductor?



First we find the equivalent inductance of the circuit reducing the circuit in steps. The two 25 H inductors are in series, so



The 30 H and 50 H inductors are in parallel, so 1/L_eq = 1/(50 H) + 1/(30 H) = 1/(18.75 H).



The time constant for the circuit is τ = L_eq/R = (18.75 H)/(100 Ω) = 0.1875 s.

The behaviour of a charging RL circuit is I(t) = (ε/R)(1 - e^-t/τ ). We are looking for t when I = 0.85I_max = 0.85ε/R. So we have

0.85 = 1 - e^-t/τ ,

which yields t = -τ ln(0.15) = 0.356 s.

The energy in the equivalent circuit is U = 1/2L_eqI². The current is

I = 0.85 × ε/R = 0.85 × (10 V)/(100 Ω) = 0.085 A.

The energy is thus

U = 1/2(18.75 H)(0.085 A)² = 67.7 mJ.

To find the energy in each individual inductor, we need to know the current in each. The 18.75 H inductor is actually a 30 H and a 50 H inductor in parallel.

We know two things about inductors in parallel, first the current through the 50 H inductor and the current through the 30 H inductor add up to the current entering:

I₃₀ + I₅₀ = I .

Second, the voltage drop is the same across both inductors

-L₃₀dI₃₀/dt = -L₅₀dI₅₀/dt .

Using both facts together we find

dI₃₀/dt = (50/80) dI/dt

and

dI₅₀/dt = (30/80) dI/dt

From this, we conclude I₃₀ = (3/8)I and I₅₀ = (5/8)I. Thus the energies in each are

U₃₀ = 1/2L₃₀I₃₀² = 1/2(30 H)[(5/8)(0.085 A)]² = 42.3 mJ ,

and

U₅₀ = 1/2L₅₀I₅₀² = 1/2(50 H)[(3/8)(0.085 A)]² = 25.4 mJ .

Note that the total energy equals 67.7 mJ.

Finally we note that the 50 H inductor is two identical 25 H inductors in series, which care the same current as the 50 H inductor. Therefore

U₂₅ = 1/2L₂₅I₂₅² = 1/2(25 H)[(3/8)(0.085 A)]² = 12.7 mJ .



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16. The inductor in the diagram below is ideal with no resistance. (a) Determine the current through the resistors, the instant that the switch S is closed. (b) Determine the currents through the resistors when the switch S has been closed for a long time. (c) Determine the time dependence of the currents. Use Kirchhoff's rules and use MAPLE so solve the resulting system of differential equations. (d) The switch S is now closed. How long will it take for the current through R₂ to drop to 20 % of its initial value?



1. Initially, the inductor acts like an open circuit so that R₁ and R₂ are in series. The circuit's behaviour is identical to

   

   Thus the current through the series combination is

   I = ε/(R₁ + R₂) .

2. After a long time the inductor acts as a short circuit bypassing R₂. The circuit is now identical to

   

   Thus the current through the series combination is

   I = ε/R₁ .

3. First we assign a current to each branch and then apply Kirchhoff's Rules.

   

   Our equations are

   I1 = I2 + IL	(1)
   ε- I1R1 - I2R2 = 0	(2)
   ε- I1R1 - LdIL/dt = 0	(3)

   Our initial condition is I_L(0) = 0, since the inductor tries to keep currents from changing. Making use of MAPLE, the solutions are:

   

   In the limit t ® 0, these become I₁(0) = I₂(0) = ε/(R₁ + R₂) and I_L(0) = 0 as we have seen in part (a). In the limit t ® ¥ , these become I₁(0) = I_L(0) = ε/R₁ and I₂(0) = 0 as we have seen in part (b).

4. We are looking the time such that I₂ = 0.2I₂(0) = 0.2ε/(R₁ + R₂). Using the solution supplied by MAPLE, we have

0.2 = e^-t(R1||R2)/L .

Solving we find

t = -ln(0.2) L/(R₁||R₂) = 1.609 L/(R₁||R₂) .

Questions? mike.coombes@kwantlen.ca