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.

Answer 2

Instead of quoting the standard energy formulas \(E_C = \tfrac12 CV^2\) and \(E_{source}=CV^2\) directly, both can be obtained by integrating instantaneous power over the charging transient and then compared.

During charging, the capacitor voltage is \(v_C(t) = V(1-e^{-t/RC})\) and the charging current is \(i(t) = \dfrac{V}{R}e^{-t/RC}\). The energy delivered by the source over all time is \[ E_{source} = \int_0^\infty V i(t)\,dt = \int_0^\infty \frac{V^2}{R}e^{-t/RC}\,dt = \frac{V^2}{R}\times RC = CV^2 \] The energy stored in the capacitor at steady state, found by integrating \(v_C i\,dt\), works out to \[ E_C = \int_0^\infty v_C(t) i(t)\,dt = \frac12 CV^2 \]

1. 0.362: This value is close to \(1/e\), which characterises the fraction of charging remaining after one time constant, not the energy ratio at steady state.
2. 0.500: Dividing the two integrals above, \(E_C/E_{source} = \tfrac{\frac12CV^2}{CV^2} = 0.5\), matching this option exactly.
3. 0.632: This value is \(1-1/e\), the fraction of full charge reached after one time constant, not an energy ratio.
4. 1.000: This would mean no energy at all is lost in the resistor during charging, which contradicts the resistor dissipating exactly as much energy as ends up stored in the capacitor.

Therefore, the correct answer is 0.500.