Question

A capacitor of 20 uF and charged to 500 V:
connected in parallel to another capacitor of
charged to 200 V. The common potential
(a) 200 V (b) 400 V (C) 800 V (d) 160​

Answer 1

Mistake in the question :

A capacitor of 20 uF and charged to 500 V and is connected in parallel to another capacitor of  10uF charged to 200 V. The common potential will be :

(a) 200 V (b) 400 V (C) 800 V (d) 160 V

Answer :

Given ,

$C_{1}$ = $20*10^{-6}F$

$C_{2}$ = $10*10^{-6}F$

$V_{1}$= 500 V

$V_{2}$ = 200 V

We know,

If two conductors are touched mutually and then separated , then the charges on them will be divided in the ratio of their capacitance.

$Q_{1} = C_{1}V_{1} \\\\Q_{2} = C_{2}V_{2}$

And, Total charge = $Q_{1}+Q_{2}$

Total capacitance = $C_{1} +C_{2}$

And the common potential = $\frac{total charge}{total capacitance}$

= $\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}$

on putting the values,

= $\frac{20*10^{-6}*500+10*10^{-6}*200}{20*10^{-6}+10*10^{-6}}$

= $\frac{10^{-6}(20*500+10*200)}{10^{-6}(20+10)}$

= $\frac{12000}{30}$

= $400 V$

Thus, Common potential is 400 V.

Answer 2

Answer:

this is your answer thanku