Consider the depicted circuit. The voltage source is a 5 V power supply. The sol
ID: 2113875 • Letter: C
Question
Consider the depicted circuit. The voltage source is a 5 V power supply. The solenoid is considered to be ideal so that it has no internal resistance. Both resistors are the same: 5.0 %u03A9.
Assume the switch had been closed for a long time. Use Kirchhoff's rules to determine the voltage drop across R2.
V2 =
At the instant just after the switch was opened, what is the current though the solenoid?
IL =
At the instant just after the switch was opened, what is the current though R2?
I2 =
At the instant just after the switch was opened, what the current though R1?
I1 =
At the instant just after the switch was opened, what the magnitude of the voltage drop across R1?
V1 =
Suppose R1 were replaced with a 5 M%u03A9 resistor and the switch had been closed for a long time. What would the magnitude of the voltage drop across R1 be just after the instant in which the switch was opened?
V1(5 M Omega) =
I said 5V, 1A, 1A, 1A, 5V, and 5e6 V
Explanation / Answer
a) Voltage across R2 (V2) will be 5v as per KVL ( 5v = CURRENT*R2)
b) At the instant switch is opened inductor will try to oppose change of current. Current at t<0 is 5/5=1A. Hence current just after switch is opened will be IL=1A
c) I2 will be = IL = 1A
d) I1 will be = IL = 1A
e) Voltage drop across R1=V1=I1*R1=1*5=5v
f)When R1 is 5 M ohm current in inductor after switch opened is also 1Amp
Voltage across R1 = 1*5*10^6 = 5 Mega volt