Question

Question #b0ae0

Answer 1

The answer will be c. 7.2V

This is very simple to solve:

We have to subtract 40% of P.D from Capacitor. Since it is charged with 12V:

40% will become 0.4 and we will multiply it by 12 giving us 4.8V.

Now, we know that the remaining P.D inside the Capacitor is 7.2V.

You can also do it by taking directly 60% into consideration:

60% will become 0.6 and we will multiply it by 12 we will get 7.2V.

Answer 2

A capacitor is charged to potential difference of 12V. It delivers 40% of its stored energy to lamp.The final potential difference across capacitor is 9.295volts.

Explanation:

Initially:
We know that
Energy stored in capacitor=`E_i=1/2CV^2`
Energy stored in capacitor=`E_i=1/2C(12)^2` [since initial voltage=12 v]
Energy stored in capacitor=`E_i=72C` joulse
After consumption of 40% energy:
Remaining energy in capacitor is 60% of initial energy.
`E_f=0.6*E_i` joulse
`E_f=0.6*(72C)` joulse
`E_f=43.2C` joulse__(1)
but `E_f=1/2C(V_f)^2`___(2)
From equation(1) and equation(2);
`1/2C(V_f)^2=43.2C`
`1/2cancel(C)(V_f)^2=43.2cancel(C)`
`V_f=9.2951` volts

The final potential difference across capacitor if capacitance is constant(didn't mentioned in question) is 9.295volts.