Question

The capacity of a condenser is 4*10^-6 f and its potential is 100 V. The energy released on discharging it fully will be ____________

[VMMC 1995]

A capacitor of 20 μF is charged up to 500 V is connected in parallel with another capacitor 10μF which is charged up to 200V. The common potential is ____________

[DPMT 1995

Answer 1

Given,

C = $4*10^{-6} F$

V = 400 V

As we know,

Heat produced = Energy released

         

                           = $\frac{1}{2}CV^{2}$

On putting the values,

= $\frac{1}{2}*4*10^{-6} *(400)^{2}$

= $= 2*160000*10^{-6} \\\\= 32*10^{-2} \\\\= 0.32 J$

Given,

C1 = 20 μF

C2 = 10 μF

V1 = 500V

V2 = 200 V

As we know,

Common potential = V = $\frac{C1V1+C2V2}{C1+C2}$

on putting the values,

= $\frac{20*500+10*200}{20+10}$

=  $\frac{12000}{30} V$

= 400V

Answer 2

Question1:- The capacity of a condenser is 4*10^-6 f and its potential is 100 V. The energy released on discharging it fully will be ____________

Solution:-

The energy released on discharging it fully, is given by

E = (1/2)CV²

E = (1/2) × 4 × 10^-6 × (100)²

E = 2 × 10^-2

E = 0.02Joules

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Question2:- A capacitor of 20 μF is charged up to 500 V is connected in parallel with another capacitor 10μF which is charged up to 200V. The common potential is ____________

Solution:-

Given:-

Capacitor C1 = 20 × 10^-6

Capacitor C2 = 10 × 10^-6

V1 = 500 V

V2 = 200 V

The common potential is given by

V = $\sf\dfrac{C1V1 + C2V2}{C1 + C2}$

V = $\sf\dfrac{20 × 10^{-6} × 500 + 10 × 10^{-6} × 200}{20 × 10^{-6} + 10 × 10^{-6}}$

After solving this we will get,

V = 400 volts

_______________________________

Hence, the common potential is 400 volts.

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