the capacitor had zero influence on the current at the moment that the function

Question

the capacitor had zero influence on the current at the moment that the function generator turned on, however the longer the function generator was turned on, however the longer the function generator was turned on the LARGER the effect the capacitor had on the current. Why is this so? the inductor (specifically the emf, epsilon) completely controlled the current at the moment that the function generator was turned on, however the longer the function generator was left on the LESS the effect the emf has on the current. Why is this so?

Explanation / Answer

2)

as the current through a charging capacitor is

I = I0 * e^(-t/T)

t is the time

hence , at t = 0 s

I = I0

from the current equation , the current in the capacitor will decrease exponentially .

thereforce , the current will be larger when the current at the moment generator is closed.

3)

for the current in the inductor is given as

I = I0 * (1 - e^(-t/T))

as t = 0

I = 0 A

hence , the longer the function generator was LESS the effect the emf has on the current

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