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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minPYQ_2017

PhysicsMediumsingle choice

Two capacitors when connected in series have a capacitance of $3 μ F$ , and when connected in parallel have a capacitance of $16 μ F$ . Their individual capacities are

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$C_{s} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = 3 C_{p} = C_{1} + C_{2} = 16 ∴ C_{1} C_{2} = 48 C_{1} - C_{2} = (C_{1} + C_{2})^{2} - 4 C_{1} C_{2} = 1 6^{2} - 4 \times 48 = 64 = 8 C_{1} + C_{2} = 16 μ F C_{1} - C_{2} = 8 μ F \Rightarrow 2 C_{1} = 24 μ F \Rightarrow C_{1} = 12 μ F ∴ C_{2} = \frac{48}{12} = 4 μ F$

Stream:VITEEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2017

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