What is the energy dissipated in the resistor R in the circuit below, for fully charging the capacitor?

This question was previously asked in

VSSC (ISRO) Technician B (Electronic Mechanic) Previous Year Paper (Held on 25 Sept 2016)

Option 2 : 1/2 CV^{2}

Subject Test 1: Electrical Basics

3098

20 Questions
80 Marks
30 Mins

__Concept__**:**

The energy dissipated by R is given by:

E = power × time

E = I^{2}Rt

\(E = \mathop \smallint \nolimits_0^t {I^2}R\;dt\)

Also, in Laplace domain, the capacitive reactance is given by:

\({X_c} = \frac{1}{{sC}}\)

__Calculation__**:**

Given circuit:

\(I\left( s \right) = \;\frac{{\frac{V}{s}}}{{R + \frac{1}{{sc}}}} = \frac{{VC}}{{sRC + 1}}\)

\(I\left( s \right) = \frac{{\frac{V}{R}}}{{s + \frac{1}{{RC}}}}\)

Taking the Laplace inverse of I(s), we get:

\(i\left( t \right) = \frac{V}{R}{e^{ - \frac{t}{{RC}}}}u\left( t \right)\)

Energy E will be:

\(E = \mathop \smallint \nolimits_0^t {I^2}R\;dt = \frac{{{V^2}}}{{{R^2}}}R\;\mathop \smallint \nolimits_0^t {e^{ - \frac{{2t}}{{RC}}}}dt\)

= \(\frac{{{V^2}}}{R}\left[ {\frac{{ - {e^{ - \frac{{2t}}{{RC}}}}}}{{\left( {\frac{2}{{RC}}} \right)}}} \right]_0^t\)

= \(\frac{1}{2}C{V^2}\left[ {1 - {e^{ - \frac{{2t}}{{RC}}}}} \right]\)

The capacitor will be fully charged for t = ∞ (steady-state), i.e. at t = ∞, the energy stored will be:

\(Energy\;\left( E \right) = \frac{1}{2}C{V^2}\)