Question

A series RL circuit consists of a resistor $R=10 \ \Omega$ and an inductor $L=1\text{ H}$, connected to a DC voltage source of $V=20\text{ V}$. The switch is closed at $t=0$, and the initial current is zero. Determine the expression for the current $i(t)$ through the inductor for $t>0$, assuming ideal components.

• A. $i(t)=2e^{-10t}\text{ A}$
• B. $i(t)=2(1-e^{-0.1t})\text{ A}$
• C. $i(t)=2(1-e^{-10t})\text{ A}$
• D. $i(t)=2e^{-0.1t}\text{ A}$

Hint:

For any RL transient problem, always identify the time constant \(\tau = L/R\) and the steady-state value \(I_{\text{final}} = V/R\) first.
This quickly eliminates incorrect options. Here, \(\tau = 0.1 \implies 1/\tau = 10\), and \(I_{\text{final}} = 2\text{ A}\), which immediately points to option (C).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the transient current expression \(i(t)\) in a series RL circuit when connected to a constant DC voltage source \(V = 20\text{ V}\) at \(t=0\), with zero initial inductor current.

Step 2: Key Formula or Approach:
The current response of a first-order RL circuit with a DC input is given by the standard transient response equation:
\[ i(t) = I_{\text{final}} + (I_{\text{initial}} - I_{\text{final}}) e^{-t/\tau} \] where:
\(I_{\text{initial}}\) is the initial current through the inductor at \(t = 0^+\),
\(I_{\text{final}}\) is the steady-state current as \(t \to \infty\), and
\(\tau\) is the time constant of the RL circuit, defined as \(\tau = \frac{L}{R}\).

Step 3: Detailed Explanation:

• Determine the initial current:
Since the switch is closed at \(t=0\) and the initial current is given as zero, we have:
\[ I_{\text{initial}} = i(0^+) = 0\text{ A} \]
• Determine the final (steady-state) current as \(t \to \infty\):
In steady-state under a DC source, the inductor behaves as a short circuit. The steady-state current is limited only by the resistor:
\[ I_{\text{final}} = \frac{V}{R} = \frac{20\text{ V}}{10\ \Omega} = 2\text{ A} \]
• Determine the time constant \(\tau\):
\[ \tau = \frac{L}{R} = \frac{1\text{ H}}{10\ \Omega} = 0.1\text{ s} \]
• Find the reciprocal of the time constant:
\[ \frac{1}{\tau} = \frac{1}{0.1} = 10\text{ s}^{-1} \]
• Substitute these values into the standard transient response equation:
\[ i(t) = 2 + (0 - 2) e^{-10t} = 2(1 - e^{-10t})\text{ A} \]

Step 4: Final Answer:
The expression for the current is \(i(t) = 2(1 - e^{-10t})\text{ A}\).