Consider the depicted circuit. The voltage source is a 5 V power supply. The sol
ID: 2185068 • 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 Ohm. Assume the switch had been closed for a long time. Use Kirchhoff's rules to determine the voltage drop across 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 MOhm 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 Ohm ) =Explanation / Answer
by applying krichoff's law v = iR1 + iR2 + iL where is L is solenoid so v across r2 is v = v2 + vL since RL is 0 so VL = iL * RL so VL = 0 so V2 = V when switch is opened v = iR1 + iR2 + iL since i is same as series cct. so iR1 = iR2 = iL iL = v - iR1 - iR2 i1 will be equal to i2 v1 = i1 * R1