Discussion Forum



1)

Two capacitors C1and C2 in a circuit are joined as shown in figure. The potentials of points A and B are V1 and V2 respectively. Then the potential of point D will be

2052021672_202332_35d9faf105da4f8ebf187f245800f83b.png


A) $\frac{V_{1}+V_{2}}{2}$

B) $\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}$

C) $\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}$

D) $\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}$

Answer:

Option C

Explanation:

Consider the potential at D be 'V'.

Potential drop across C1 is (V - V1) and C2 is (V - V)

.'. q1 = C1 (V - V1), q2 = C2 (V2 - V)

As q1 = q2 [capacitors are in series]

.'. C1 (V - V1) = C2 (V2 - V) 

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