Question

The current through a series RL circuit is \( \frac{1}{4} e^{-\frac{3}{4} t} \) when excited by a unit impulse voltage. The values of R and L are respectively

• A. 8, 4
• B. 4, 2
• C. 2, 4
• D. 1, 4

Hint:

For a series RL circuit, use the formula \( \tau = \frac{L}{R} \) to calculate the time constant and determine the values of \( R \) and \( L \).

Answer 1

The Correct Option is C

The given current equation represents the transient response of a series RL circuit to a unit impulse voltage. The general form of the transient response for an RL circuit is: \[ I(t) = \frac{1}{\tau} e^{-\frac{t}{\tau}} \] where \( \tau = \frac{L}{R} \) is the time constant of the circuit.
- In the given equation, the time constant is represented by \( \tau = \frac{4}{3} \).
- From the equation, \( I(t) = \frac{1}{4} e^{-\frac{3}{4} t} \), we can compare and determine that: \[ \tau = \frac{L}{R} = \frac{4}{3} \] - Solving for \( L \) and \( R \), we find that the values of \( R = 2 \, \Omega \) and \( L = 4 \, H \).
Thus, the correct answer is option (3).