A circuit is constructed with four resistors, one inductor, one battery and a sw
ID: 1400505 • Letter: A
Question
A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 42 , R3 = 72 and R4 = 148 . The inductance is L = 206 mH and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.
1)
The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?
A
2)
What is I4(), the magnitude of the current through the resistor R4 after the switch has been closed for a very long time?
A
3)
What is IL(), the magnitude of the current through the inductor after the switch has been closed for a very long time?
A
4)
After the switch has been closed for a very long time, it is then opened. What is I3(topen), the current through the resistor R3 at a time topen = 3.6 ms after the switch was opened? The positive direction for the current is indicated in the figure.
A
5)
What is VL,max(closed), the magnitude of the maximum voltage across the inductor during the time when the switch is closed?
V
6)
What is VL,max(open), the magnitude of the maximum voltage across the inductor during the time when the switch is open?
Explanation / Answer
(1) At the time of closing of switch , inductor behave as open circuit therefore
net resistance = R1 + R3 + R4 = 42+72+148 = 262 ohm
Current = V/R = 12/262 = 0.0458 A
(2) In this case we can ignore the reactance of the inductor due to negligible value
therefore R2 and R3 will be in the parallel
so they will give 26.5263 ohm
and it will be in the series with R1 and R4
so net resistance = 26.5263+42+148 = 216.5263 ohm
So I = V/R = 12/216.5263 = 0.0554 A
(3) Current in inductor
potential drop = 12 - 0.0554(42+148) = 1.474
IL = 1.474/42 = 0.035 A
(4) when switch is open then R1 and R4 will be no more in the ckt.
i = IL (1- e-(R/L)t)
t = 3.6 ms R = R2+ R3 = 42+72 = 114 , L = 206 mH
so putting these values we get
I = 0.030226 A
Sorry we are allowed to solve four sub parts only.
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