After a long time, the switch in the above circuit is opened, abruptly disconnec

Question

After a long time, the switch in the above circuit is opened, abruptly disconnecting the battery from the circuit. What is the current I through the vertical resistor immediately after the switch is opened? (+ is in the direction of the arrow)
(A) I = V/R
(B) I = V/2R
(C) I = 0
(D) I = -V/2R
(E) I = -V/R

I know the answer is E. I need to know why it's negative though. Wouldn't it simply be I =V/R?

Assume the switch had been closed for a long time before opening.

After a long time, the switch in the above circuit is opened, abruptly disconnecting the battery from the circuit. What is the current I through the vertical resistor immediately after the switch is opened? (+ is in the direction of the arrow) I = V/R I = V/2R I = 0 I = -V/2R (E) I = -V/R

Explanation / Answer

when the circuit was intiially closed , no current was flowing in the circuit containing the vertical resistance

sio the effective current in the circuit was I=V/R

now when the circuit is opened,

current in the inductor=V/R (in downward direction)

hence by KCL,

current in the ressitor flows up hence it is negative

so I=-V/.R

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