Question

Find the voltage across the capacitor in steady state. 

• A. \(1\,\text{V}\)
• B. \(0.5\,\text{V}\)
• C. \(\dfrac{3}{2}\,\text{V}\)
• D. \(4\,\text{V}\)

Hint:

In steady state DC circuits: \[ \text{Capacitor} \rightarrow \text{Open circuit} \] \[ \text{Inductor} \rightarrow \text{Short circuit} \] This greatly simplifies circuit analysis.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a DC circuit, when the steady state is reached, a capacitor becomes fully charged and behaves like an open circuit.
This means no direct current flows through the branch containing the capacitor.
The circuit then acts like a simple resistive loop for the flow of steady current.
Step 2: Key Formula or Approach:
Identify the steady-state path of the current, compute total equivalent resistance $R_{eq}$, find total current $I = \frac{V_{\text{battery}}}{R_{eq}}$, and then determine the voltage drop across the elements parallel to the capacitor.
Step 3: Detailed Explanation:
In steady state, the current through the middle branch (containing the capacitor and 4$\Omega$ resistor) is zero.
The active current loop will only consist of the 2V battery, the 6$\Omega$ resistor, and the top 2$\Omega$ resistor.
The total resistance of this series loop is:
\[ R_{eq} = 2\Omega \ (\text{top branch}) + 6\Omega \ (\text{bottom branch}) = 8\Omega \]
The steady current $I$ flowing through this active loop is:
\[ I = \frac{V}{R_{eq}} = \frac{2\text{ V}}{8\Omega} = 0.25\text{ A} \]
The voltage across the top branch (2$\Omega$ resistor) is:
\[ V_{2\Omega} = I \times 2\Omega = 0.25\text{ A} \times 2\Omega = 0.5\text{ V} \]
Since the middle branch is in parallel with the top branch, the total voltage across the entire middle branch is also $0.5\text{ V}$.
Because there is zero current flowing through the middle branch, there is zero voltage drop across the 4$\Omega$ resistor ($V_{4\Omega} = 0 \times 4 = 0$).
Thus, the entire voltage of the branch appears across the capacitor:
\[ V_C = V_{2\Omega} - V_{4\Omega} = 0.5\text{ V} - 0\text{ V} = 0.5\text{ V} \]
Step 4: Final Answer:
The voltage across the capacitor in steady state is 0.5V.