Question

If $C_1=14\mu \text{F}$ and $C_2=38\mu \text{F}$, what is the charge on $C_2$ in the steady state?

• A. $5500\mu \text{C}$
  
• B. $0$
• C. $4160\mu \text{C}$
• D. $2080\mu \text{C}$

Answer 1

The correct option is B $0$

Due to the presence of conducting wire $\text{AB}$, no charge will be accumulated on the plates of the capacitors.

$\text{AB}$ path is a short circuit path, so all the charges bypass through $\text{AB}$.

Hence, the charges on both the capacitors will be $0$

Why this Question ? Key point: Whenever a short-circuit path is introduced in the direction of charge flow, entire charge passes through the short-circuit path. These type of circuits are also called as bypass circuits.