I have tried to solve these, but those solutions on the paper might be uncorrect
ID: 1426786 • Letter: I
Question
I have tried to solve these, but those solutions on the paper might be uncorrect.
Postlab H7 Induction 1) In the circuit shown, the voltage source is a 5 V power supply. The inductor is considered to be ideal (it has no internal resistance) and large (say, 1 henry), R1 =1 and R2-3 Units are required in all numerical answers Assume the switch had been closed for a long time. Use Kirchhoffs rules to determine the voltage drop across R2 R. At the instant just after the switch was opened, what is the current though the inductor? At the instant just after the switch was opened, what is the current though R2? At the instant just after the switch was opened, what the current though R,? At the instant just after the switch was opened, what the magnitude of the voltage drop across R? Suppose R1 were replaced with a 10 resistor and the switch had been closed for a long time. What would the magnitude of the voltage drop across R, be just after the instant in which the switch was opened? With R1 = 10 , VI =Explanation / Answer
1) when the switch is closed for a long time, the two resistors are in parallel combination and the drop across inductor will be zero [ as per, V = Vb e-tR/L , t = infinite ]
The drop across resistors would be equal to 5 volts
Hence, drop across R2 = 5 volts
2)Just after the switch opens, current in the inductor is zero.
3)When the switch opens, the two resistors are in series, so the same current will flow through them,
Req = 1 + 3 = 4 ohm,
current in the circuit from Ohms law,
I = V/R = 5/3 = 1.25 A
Hence, current through R2 = I2 = 1.25 A
4) the same amount of current will flow through R1 also as it is in series with R2, so
I1 = 1.25 A.
5) the drop across R1 will be
V1 = I1 R1 = 1.25 x 1 = 1.25 Volts
6) in that case, Req = 10 + 3 = 13 ohm
I1 = I2 = 5/13 = 0.385 A
V1 = I1 x R1 = 0.385 x 10 = 3.85 Volts