Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
Advertisement Remove all ads

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.

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

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

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×