JEE Physics 2023 Question with Solution

Given: Three capacitors are connected in parallel with values $C_1 = 2\,\mu F$, $C_2 = x\,\mu F$, and $C_3 = 3\,\mu F$. The applied potential difference is $V = 10\,V$ and the total charge stored is $Q = 100\,\mu C$.

Find: The value of $x$.

For capacitors in parallel,

$$C_{eq} = C_1 + C_2 + C_3 = 2 + x + 3 = (x+5)\,\mu F$$

Using the relation between charge, capacitance and voltage,

$$Q = C_{eq}V$$

Substitute the given values,

$$100 = (x+5)\times 10$$ $$x + 5 = 10$$ $$x = 5$$

Therefore, the value of $x$ is $5$.

Given: The capacitors are in parallel, so their capacitances add directly.

Find: The value of $x$.

First find the equivalent capacitance from $Q = C_{eq}V$:

$$C_{eq} = \frac{Q}{V} = \frac{100\,\mu C}{10\,V} = 10\,\mu F$$

Now use

$$2 + x + 3 = 10$$ $$x = 5$$

This works because in a parallel combination, each capacitor has the same potential difference and the total capacitance is the sum of individual capacitances. Hence, the value of $x$ is $5$.