Question

A capacitor of capacity C1 is charged upto V volt and then connected to an uncharged capacitor of capacity C2. The final potential difference across each will be

1- C2V/C1+C2

2- C1V/C1+C2

3- (1+C2/C1)V

4-(1-C2/C1)V

Answer 1

Answer:

C1V/C1 + C2

Explanation:

The formula for common potential difference

V' = C1V1 + C2V2

       C1 + C2

As the second capacitor is uncharged

Q2= C2V2=0

and the first capacitor is charged upto V volts,

Q1=C1V1 =C1V

   Therefore, V'= C1V + 0

                            C1+ C2

             V'= C1V

                 C1+ C2

Answer 2

Answer:

The final potential difference across each will be C₁V/C₁+C₂ i.e. option (2).

Explanation:

The charge on the first capacitor is given as,

$Q_1=C_1V$            (1)

Where,

Q₁=charge on the first capacitor

C₁=capacity of the first capacitor

V=potential difference across the first capacitor

According to the question, the capacitor having capacity C₂ is uncharged. So,

$Q_2=0$           (2)

When the capacitors are connected in parallel,

$C_e_q_v=C_1+C_2$           (3)

$Q_e_q_v=Q_1+Q_2$         (4)

Let V' be the final potential across each capacitor,

$V'=\frac{Q_e_q_v}{C_e_q_v}$                (5)

By substituting equations (3) and (4) in equation (5) we get;

$V'=\frac{Q_1+Q_2}{C_1+C_2}$          (6)

By substituting equations (1) and (2) in equation (6) we get;

$V'=\frac{C_1V+0}{C_1+C_2}$

$V'=\frac{C_1V}{C_1+C_2}$

Hence, the final potential difference across each will be C₁V/C₁+C₂ i.e. option (2).