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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minPYQ_2018

PhysicsMediumsingle choice

A fully charged parallel plate capacitor of $1 μ F$ capacity is discharging through a resistor. If the energy of the capacitor reduces to half in one second then the value of resistance will be

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Solution:

As $Q = Q_{0} e^{- t / τ}$ . When energy is $50%$ then $Q = \frac{Q _{0}}{2}$ $\frac{Q _{0}}{2} = Q_{0} e^{- t / τ} e^{t / τ} = 2 \frac{t}{τ} = ln (2) τ = \frac{t}{ln ( 2 )} RC = \frac{2}{ln 2} R = \frac{2}{C ln 2} = \frac{2}{1 0 ^{- 6} \times ln 2} = \frac{2 \times 1 0 ^{6}}{ln 2} Ω$

Stream:AIIMSSubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2018

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