Question

The equivalent capacitance in the circuit between $\text{A}$ and $\text{B}$ is

• A. $1\mu \text{F}$
• B. $2\mu \text{F}$
• C. $3\mu \text{F}$
• D. $1/3\mu \text{F}$

Answer 1

The correct option is C $3\mu \text{F}$

To understand it better, let's connect a battery of emf $E$ across $\text{A}$ and $\text{B}$.


Since, we know that potential drop across the conducting wire is zero means potential throughout the wire remains same.

$\therefore V_Q=V_B=V_R=V_S=0\text{V}$

And, $V_P=V_A=V_O=V_T=E$

$\Rightarrow$ We can clearly see that potential difference across all three capacitors is $E$ means all are in parallel combination. Therefore, the circuit can be reduced as shown.

$\therefore C_{eq}=(1+1+1)\mu \text{F}=3\mu \text{F}$

Hence, option $\left(c\right)$ is correct.

Key concept: The voltage across capacitors in parallel are equal.