Consider the circuit shown in the figure. The battery has emf ? = 75 volts and n
ID: 2257277 • Letter: C
Question
Consider the circuit shown in the figure. The battery has emf ? = 75 volts and negligible internal resistance. The inductance is L = 0.5 H and the resistances are R 1 = 12 ? and R 2 = 9.0 ?. Initially the switch S is open and no currents flow. Then the switch is closed.
What is the current in the resistor R 1 just after the switch is closed?
After leaving the switch closed for a very long time, it is opened again. Just after it is opened, what is the current in R 1?
Consider the circuit shown in the figure. The battery has emf ? = 75 volts and negligible internal resistance. The inductance is L = 0.5 H and the resistances are R 1 = 12 ? and R 2 = 9.0 ?. Initially the switch S is open and no currents flow. Then the switch is closed. What is the current in the resistor R 1 just after the switch is closed? After leaving the switch closed for a very long time, it is opened again. Just after it is opened, what is the current in R 1?Explanation / Answer
just after the swith is closed no current passes throught the branch containing R2 and L.. beacuse inductor acts as an open circuit immediately after closing the switch.
current in R1 = 75 / 12 A = 6.045 A
After leaving the switch closed for sufficiently long time the inductor reaches steady state and in steady state inductor acts as a short circuit and offers zero resistance effectively. It is as if only 2 resistors are connected in parallel.
In the steady state current across inductor = current across R2
current across R2 = 75 / 9 = 8.33 A
Now after opening the switch the current in the inductor branch doesn't change immeditely because inductor resists change. The current flowing in R1 = current flowing in inductor branch = 75 / 9 A