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: Total energy delivered by the source while charging: \(E_{source}=CV^2\).
Step 2: Energy stored in the capacitor once fully charged: \(E_C = \dfrac12 CV^2\).
Step 3: Take the ratio: \(\dfrac{E_C}{E_{source}} = \dfrac{\frac12CV^2}{CV^2}\).
\[ \boxed{\text{Ratio} = 0.5} \]

Answer 2

A neat way to see why the ratio is exactly \(0.5\) regardless of the resistor's value is to note that the resistor's job during charging is entirely to dissipate the difference between what the source supplies and what ends up stored, and this split does not depend on how large \(R\) is, only on the fact that the capacitor's voltage rises exponentially toward \(V\).

Since \(E_{source}=CV^2\) and \(E_C=\tfrac12CV^2\) regardless of \(R\) (a larger \(R\) only slows down the charging, it does not change the final energies), the energy dissipated in the resistor is \[ E_R = E_{source}-E_C = CV^2 - \tfrac12CV^2 = \tfrac12CV^2 \] which is exactly equal to \(E_C\) itself. In other words, charging a capacitor through any resistor always splits the source's energy exactly in half between the capacitor and the resistor.

1. 0.362: Does not correspond to an equal split between capacitor and resistor.
2. 0.500: Matches the equal 50-50 split between \(E_C\) and \(E_R\) found above.
3. 0.632: Also does not correspond to an equal energy split.
4. 1.000: Would mean the resistor dissipates nothing, which is inconsistent with \(E_R=E_C=\tfrac12CV^2\) both being nonzero.

Therefore, the correct answer is 0.500.