Physics 302/328 Homework
Chapter 6 – First-Order Circuits
‡ Problem 1: Use the step-by-step method to find i0 (t) for t > 0 in the circuit
below.
Solution:
Before the switch is closed, the 12V source produces a 6V potential across the 6kW resistor,
causing a current of i = 1.0 mA at t = 0 - . Since there is no current through the 2 k resistor
at this time, there is no voltage drop and both sides of the resistor are at the same potential.
Thus, the voltage across the 200 µF capacitor is 6.0 Volts at at t = 0 - .
When the switch is closed, the contribution from the voltage source is shorted out and the
capacitor remains charged at 6.0 Volts at t = 0 + . The current flowing out of the capacitor is
3k
6V / (2kW + 3kW//6kW) = 1.5 mA. Using current division, we get i0 (0 + ) = 1.5 mA ( 3 k -6 k ) =
0.5 mA.
After a long time i0 Ø 0, or i0 (¶) = 0.
The Thevenin resistance seen by the capacitor is (2kW + 3kW//6kW) = 4 kW, so
t = RTh C = 4000W (200 µW) = 0.8 sec.
The solution is i0 (t) = i0 (¶) + [i0 (0 + ) - i0 (¶)] ‰ -têt
i0 (t) = 0 + [0.5 mA - 0] ‰ -tê0.8 = 0.5 mA ‰ -tê0.8
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Chapter 7.11.nb
In[2]:=
PlotA0.5 ‰-tê0.8 , 8t, 0, 5<, PlotRange Ø 80, 0.5<E
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‡ Problem 2:
Find v0 (t) for t > 0 in the circuit below, using the step-by-step method. (corrected)
Solution:
Before the switch is thrown, acts like a short circuit, so we may ignore the 4W resistor. The
total current coming out of the 12V supply is 12 / [2W+3W//6W] = 3.0 A. The inductor current
3
can be found by current division iL H0 - L = 3.0 * 3 +6 = 1.0 A , is 1.0 A to the right.
iL H0 - L = iL H0 + L = 1.0 A .
After the switch is thrown, the inductor current begins to change and it no longer acts like a
short-circuit. That means that the 4W resistor must be included. The current through the 6W
4
resistor is found by current division: 1A 4+H6 +3L = 0.3077A and
+
Chapter 7.11.nb
Solution:
Before the switch is thrown, acts like a short circuit, so we may ignore the 4W resistor. The
total current coming out of the 12V supply is 12 / [2W+3W//6W] = 3.0 A. The inductor current
3
can be found by current division iL H0 - L = 3.0 * 3 +6 = 1.0 A , is 1.0 A to the right.
iL H0 - L = iL H0 + L = 1.0 A .
After the switch is thrown, the inductor current begins to change and it no longer acts like a
short-circuit. That means that the 4W resistor must be included. The current through the 6W
4
resistor is found by current division: 1A 4+H6 +3L = 0.3077A and
V0 H0 + L = 6.0 W * 0.3077 A = 1.846 Volt
A long time after the switch is thrown, iL H¶L = 0 , so V0 H¶L = 0
The Thevenin resistance seen by the inductor is 4W//(3W+6W) = 2.77W.
The time constant is t = L / RTh = (0.5H)/(2.77W) = 0.181 s
The solution is V0 (t) = V0 (¶) + [V0 (0 + ) - V0 (¶)] ‰ -têt
V0 (t) = 0 + [1.846V - 0] ‰ -tê0.181 = 1.846V ‰ -tê0.181
In[1]:=
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2.0
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Out[1]=
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0.0
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1.5
2.0
‡ Problem 3:
Use the step-by-step method to find v0 (t) for t > 0 in the circuit below.
3
4
Chapter 7.11.nb
Problem 3:
Use the step-by-step method to find v0 (t) for t > 0 in the circuit below.
Solution:
Initially, the capacitor voltage is equal to the voltage drop across the 6 kW resistor.
2 kW
v0 H0 - L = H6 mAL J 2 kW+6 kW N • 6 kW = 9.0 V = v0 H0 + L
When the switch opens, the capacitor is still charged to 9.0 Volts.
v0 H¶L = 0
t = R C = (6kW)(100 mF) = 0.6 s
Then v0 HtL = 9.0 V ‰ -tê0.6
In[7]:=
PlotA9 ‰-tê0.6 , 8t, 0, 4<, PlotRange Ø 80, 10.0<E
10
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Out[7]=
4
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