Maximum Rate of Energy Storage in Capacitor – Rankers Physics

Charging and Discharging of Capacitors: Practice Problem & Solution

A capacitor of capacitance \(C\) is connected to a battery of emf \(\varepsilon\) at \(t = 0\) through a resistance \(R\). Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value ?
\(\frac{\varepsilon^2}{4R}\)
\(\frac{\varepsilon^2}{2R}\)
\(RC\)
\(CR \ln 2\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Charging and Discharging of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The energy stored rate is \(P = \frac{\varepsilon^2}{R} e^{-t/RC}(1-e^{-t/RC})\). This is maximum when \(e^{-t/RC} = \frac{1}{2}\). The maximum rate is \(P_{max} = \frac{\varepsilon^2}{4R}\). This occurs at \(t = RC \ln 2\).

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