Combination of Capacitors: Practice Problem & Solution
Assertion (A): When two capacitors of capacitance \(300 \text{ pF}\) and \(600 \text{ pF}\) which can work upto maximum potential of \(4 \text{ kV}\) and \(3 \text{ kV}\) respectively, are connected in series, their combination can work upto maximum potential of \(7 \text{ kV}\). Reason (R): In series combination, maximum working potential will be sum of maximum working potential of individual capacitors.
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Combination of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:
For capacitors in series, the maximum charge \(Q_{\text{max}}\) the combination can hold is the minimum of individual \(C V_{\text{max}}\). Here, \(Q_{1, \text{max}}\ = 300 \text{ pF} \cdot 4 \text{ kV} = 1200 \text{ pC}\) and \(Q_{2, \text{max}}\ = 600 \text{ pF} \cdot 3 \text{ kV} = 1800 \text{ pC}\). So, \(Q_{\text{max}}\ = 1200 \text{ pC}\). Equivalent capacitance \(C_{\text{eq}}\ = (300 \cdot 600) / (300 + 600) = 200 \text{ pF}\). The maximum potential for the combination is \(V_{\text{max}}\ = Q_{\text{max}} / C_{\text{eq}}\ = 1200 \text{ pC} / 200 \text{ pF} = 6 \text{ kV}\). Thus, both Assertion (A) and Reason (R) are false.
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