Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# By Evaluating ∫I2rdt, Show that When a Capacitor is Charged by Connecting It to a Battery Through a Resistor, the Energy Dissipated as Heat Equals the Energy Stored in the Capacitor. - Physics

Sum

By evaluating ∫i2Rdt, show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.

#### Solution

The growth of charge on the capacitor at time t ,

$Q = Q_0 \left( 1 - e^{- \frac{t}{RC}} \right)$

$i = \frac{dQ}{dt} = \left( \frac{Q_0}{RC} \right) e^{- \frac{t}{RC}}$

Heat dissipated during time t1 to t2,

$U = \int_{t_1}^{t_2} i^2 Rdt$

$= \frac{{Q_0}^2}{2C}\left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)$

$\because Q_0 = C V_{0,}$

$U = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)$

The potential difference across a capacitor at any time t,

$V = V_0 \left( 1 - e^{- \frac{t}{RC}} \right)$

The energy stored in the capacitor at any time t,

$E = \frac{1}{2}C V^2 = \frac{1}{2}C {V_0}^2 \left( 1 - e^{- \frac{2t}{RC}} \right)^2$

∴ The energy stored in the capacitor from t1 to t2,

$E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} \right) - \frac{1}{2}C {V_0}^2 \left( 1 - e^{- \frac{2 t_2}{RC}} \right)$

$\Rightarrow E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)$

Concept: Energy Stored in a Capacitor
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 10 Electric Current in Conductors
Q 77 | Page 203