Question

In the circuit shown in Fig, capacitors $A$ and $B$ have identical geometry, but a material of dielectric constant 3 is present between the plates of $B$. The potential difference across $A$ and $B$ are respectively

• A. $2.5\text{V},7.5\text{V}$
• B. $2\text{V},8\text{V}$
• C. $7.5\text{V},2.5\text{V}$
• D. $8\text{V},2\text{V}$

Answer 1

The correct option is C $7.5\text{V},2.5\text{V}$

Let Capacitance of $A$ is $C$
Capacitance of $B$ is $KC=3C$
Now $V_A$ be potential across A, $V_B$ be potential across B.
$\Rightarrow \frac{V_A}{V_B}=\frac{3C}{C}=3$
We know that,
$V_A+V_B=10\Rightarrow 4V_B=10\Rightarrow V_B=2.5V,V_A=7.5V$