Question:
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are :
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are :
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The steeper line A is the parallel combination (\(C_1 + C_2 = 500/10\)), line B the series one (\(\frac{C_1C_2}{C_1+C_2} = 80/10\)).
Updated On: Nov 5, 2026
- \(40\,\mu\mathrm{F}\) and \(10\,\mu\mathrm{F}\)
- \(60\,\mu\mathrm{F}\) and \(40\,\mu\mathrm{F}\)
- \(20\,\mu\mathrm{F}\) and \(30\,\mu\mathrm{F}\)
- \(50\,\mu\mathrm{F}\) and \(30\,\mu\mathrm{F}\)
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The Correct Option is A
Solution and Explanation
The slope of a q-V line gives the capacitance. Line A (steeper): \(C_A = \frac{500\,\mu\mathrm{C}}{10\,\mathrm{V}} = 50\,\mu\mathrm{F}\); line B: \(C_B = \frac{80}{10} = 8\,\mu\mathrm{F}\).
The parallel combination has the larger capacitance: \(C_1 + C_2 = 50\). Series: \(\frac{C_1C_2}{C_1 + C_2} = 8 \Rightarrow C_1C_2 = 400\).
\(C_1, C_2\) are roots of \(C^2 - 50C + 400 = 0\), i.e. \((C - 40)(C - 10) = 0\). So \(40\,\mu\mathrm{F}\) and \(10\,\mu\mathrm{F}\).
Answer: (1) \(40\,\mu\mathrm{F}\) and \(10\,\mu\mathrm{F}\)
The parallel combination has the larger capacitance: \(C_1 + C_2 = 50\). Series: \(\frac{C_1C_2}{C_1 + C_2} = 8 \Rightarrow C_1C_2 = 400\).
\(C_1, C_2\) are roots of \(C^2 - 50C + 400 = 0\), i.e. \((C - 40)(C - 10) = 0\). So \(40\,\mu\mathrm{F}\) and \(10\,\mu\mathrm{F}\).
Answer: (1) \(40\,\mu\mathrm{F}\) and \(10\,\mu\mathrm{F}\)
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