Question

A \(10~\Omega\) resistor, a \(1~H\) inductor and a \(1~F\) capacitor are connected in parallel. The combination is driven by a unit step current. Under steady-state conditions, the source current flows through

• A. Resistor only
• B. Inductor only
• C. Capacitor only
• D. Resistor, Inductor and Capacitor

Hint:

Remember the steady-state DC equivalents: \[ \text{Inductor} \rightarrow \text{Short Circuit} \] \[ \text{Capacitor} \rightarrow \text{Open Circuit} \] These substitutions simplify many transient and steady-state circuit problems.

Answer 1

The Correct Option is B

Concept: The behavior of circuit elements under steady-state DC conditions is very important.
• A resistor offers finite resistance and can carry current.
• An inductor behaves as a short circuit under steady-state DC conditions.
• A capacitor behaves as an open circuit under steady-state DC conditions. Therefore, after a long time, the current distribution depends on the equivalent DC behavior of the elements.

Step 1: Determine the steady-state behavior of the inductor.
For an inductor, \[ V_L=L\frac{di}{dt}. \] Under steady-state conditions, \[ \frac{di}{dt}=0. \] Therefore, \[ V_L=0. \] Hence, the inductor behaves as a short circuit.

Step 2: Determine the steady-state behavior of the capacitor.
For a capacitor, \[ i_C=C\frac{dv}{dt}. \] At steady state, \[ \frac{dv}{dt}=0. \] Therefore, \[ i_C=0. \] Hence, the capacitor behaves as an open circuit.

Step 3: Determine the current path.
Since the inductor becomes a short circuit, the voltage across all parallel branches becomes zero. Therefore, \[ I_R=\frac{V}{R}=0. \] Also, \[ I_C=0. \] Thus the entire source current flows through the inductor branch. \[ \boxed{\text{Current flows through the inductor only}} \]