Question

. A 500 uF capacitor is charged at a stead rate of 100uC per second. A potential difference of 10V will be developed between the capacitor plates after how much time​

Answer 1

Answer:

Correct option is

D

50 sec

Given,

Capacitance, C=500μF

Potential difference, V=10V

Rate of charging, q=100μC/sec

=100×10

−6

C/sec

To fine, t=?

We know that,

Q=qt

t=

q

Q

=

100×10

−6

C

5×10

−3

C

×sec

=50sec

∴ It will take 50sec to raise the potential difference .

Hence,

Option D is correct.