Question

A DC voltage source is connected across a.series R-C circuit. when steady state reaches, the ratio of energy stored in the capacitor to the total energy supplied by the voltage source is given by
(Assume that the circuit is initially relaxed)

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $0.632$
• D. $1$

Answer 1

The correct option is A $\frac{1}{2}$

The energy stored by the capacitor inthe steady state is = $\frac{1}{2}C{V}^2$

Energy supplied by the source, $E={\int }_0^{\infty }p\left(t\right)dt={\int }_0^{\infty }ndt$

$\because i=\frac{V}{R}{e}^{\frac{-t}{v}}$

$={\int }_0^{\infty }\frac{V^2}{R}{e}^{\frac{-t}{v}}dt=C{V}^2$

The total energy supplied by the voltage source = $C{V}^2$

$\therefore$ The ratio of energy stored bythe capacitor to the energy supplied by the voltage source is $\frac{1}{2}$