CBSE 2025, Set 3

Question

In the circuit shown, the charge on the left plate of $20 \mu	ext{F}$ capacitor is $50 \mu	ext{C}$. The charge on the right plate of the $12 \mu	ext{F}$ capacitor is :

![Circuit](https://www.edvaya.com/images/cbse/physics/2025/set3/q01-capacitors-series-parallel-circuit.webp) A). $20 \mu	ext{C}$ B). $-20 \mu	ext{C}$ C). $30 \mu	ext{C}$ D). $-30 \mu	ext{C}$

Edvaya · Accepted Answer

The correct answer is $30 \mu	ext{C}$. The $12 \mu	ext{F}$ and $8 \mu	ext{F}$ capacitors are in parallel, so their equivalent capacitance is:

$$C_{p} = 12 \mu	ext{F} + 8 \mu	ext{F} = 20 \mu	ext{F}$$

This parallel combination is in series with the first $20 \mu	ext{F}$ capacitor. In a series circuit, the total charge $Q$ on each equivalent capacitor is the same. Therefore, the total charge on the parallel combination is $50 \mu	ext{C}$.

The charge divides between the $12 \mu	ext{F}$ and $8 \mu	ext{F}$ capacitors in direct proportion to their capacitances:

$$Q_{12} = \left( \frac{12}{12 + 8} ight) 	imes 50 \mu	ext{C} = \frac{12}{20} 	imes 50 \mu	ext{C} = \boxed{30 \mu	ext{C}}$$

> In parallel combinations, charge distributes proportionally to capacitance: $Q \propto C$.

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