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

Answer 1

q=CV(1-e^-t/CR)

energy stored by the capacitor=1/2CV^2
energy supplied by battery = CV^2

therefore , ratio =1/2

Answer 2

The energy stored by the capacitor in steady-state condition = $\frac{1}{2}CV^{2}$

The Energy supplied by the source (E) =∫ p(t)dt    ( limits are from 0 to ∞)

Given,

dc voltage source is connected across a series r-c circuit.

Thus, Energy supplied by the source (E) = $CV^{2}$

The ratio of energy stored in the capacitor to the total energy supplied by the voltage source = $\frac{\frac{1}{2}CV^{2} }{CV^{2} }$

                                = $\frac{1}{2}$