Question

In the circuit $C_1=20\mu F,C_2=40\mu F$ and $C_3=50\mu F$. If no capacitor can sustain more than $50V$, then maximum
potential difference between $X$ and $Y$ is

• A. 95 V
• B. 75 V
• C. 150 V
• D. 65 V

Answer 1

The correct option is A 95 V

$\text{Step 1: Voltage on each capacitor [Refer Fig.]}$

Given, $C_1=20\mu f,C_2=40\mu f,C_3=50\mu f$

Since all the capacitors are in series, so Charge on them will be same, Let $q$

Thus potential across each capacitor will be

$V_1=\frac{q}{C_1}=\frac{q}{20}$ $....\left(1\right)$

$V_2=\frac{q}{C_2}=\frac{q}{40}$ $....\left(2\right)$

$V_3=\frac{q}{C_3}=\frac{q}{50}$ $....\left(3\right)$

$\text{Step 2: Maximum voltage is equal to 50V}$

The capacitor having minimum capacitance will have maximum voltage, and this maximum voltage should not exceed $50V$

From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$

$V_1>V_2>V_3$

Here $V_1$ is maximum.

So, $V_1=50\Rightarrow \frac{q}{20}=50$

$\Rightarrow$ $q=1000\mu C$

$\text{Step 3: Potential difference across X and Y}$ $\text{[Ref. Fig. ]}$

While writing potential difference $V_{XY}$, we go from X to Y, increase in voltage is considered as positive and decrease in voltage as negative.

$V_X-\frac{q}{C_1}-\frac{q}{C_2}-\frac{q}{C_3}=V_Y$

$V_X-V_Y=\frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}=\frac{1000}{20}+\frac{1000}{40}+\frac{1000}{50}$

$V_{XY}=95volt$

Hence potential difference across $X$ and $Y$ is $95V$

Option $\left(A\right)$ correct.