Question

Consider a DC voltage source connected to a series R–C circuit. When the steady-state reaches, the ratio of the energy stored in the capacitor to the total energy supplied by the voltage source, is equal to

• A. 0.362
• B. 0.500
• C. 0.632
• D. 1.000

Hint:

In a DC R–C circuit, exactly half of the energy supplied by the source is stored in the capacitor, while the other half is dissipated in the resistor.

Answer 1

The Correct Option is B

Step 1: Energy supplied by the DC source.
When a DC voltage source of voltage $V$ is connected to a series R–C circuit, the total energy supplied by the source during charging is given by:
\[ E_{\text{source}} = CV^2 \]
Step 2: Energy stored in the capacitor at steady state.
At steady state, the capacitor is fully charged and the energy stored in it is:
\[ E_C = \frac{1}{2}CV^2 \]
Step 3: Energy dissipated in the resistor.
The remaining energy supplied by the source is dissipated as heat in the resistor. Hence, half of the supplied energy is lost in resistance.
Step 4: Ratio calculation.
\[ \text{Ratio} = \frac{E_C}{E_{\text{source}}} = \frac{\frac{1}{2}CV^2}{CV^2} = \frac{1}{2} = 0.5 \]
Step 5: Final conclusion.
Therefore, the ratio of energy stored in the capacitor to the total energy supplied by the source is 0.5.