Question

In the circuit shown the cell is ideal with emf 15 volt and each resistance is of 3 ohm the potential difference across a capacitance in the steady state is

Answer 1

Answer:

Hey mate.....

Here's ur answer ::::::'''

Given : Resistance (R) = 3 ohm

Potential difference (V) = 15 V

for steady time, t = 1 sec

We have,

«» Capacitance (C) = Q/V

But » Q = V/R*t

.°. ››› C = V / R*t*V

«»»»»» C = 1/R*t

»›‹ » C = 1/3*1

.°. »»» C = 1/3

««««»»» C = 0.3333 farad.

Hope it helps!!!

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