Question

The current through a series RL circuit, subjected to a constant emf \(\epsilon\), obeys \(L\frac{di}{dt}+iR=\epsilon\). Let L=1mH, R=1k\(\Omega\) and \(\epsilon\)=1V. The initial condition is i(0)=0. At t=1\(\mu s\), the current in mA is

• A. 1-2e^-2
• B. 1-2e^-1
• C. 1-e^-1
• D. 2-2e^-1

Answer 1

The Correct Option is C

To solve the problem of finding the current in a series RL circuit at \( t = 1 \, \mu s \), we begin with the given differential equation:

\(L\frac{di}{dt} + iR = \epsilon\)

We know the following values:

• Inductance (\( L \)) = 1 mH = \( 1 \times 10^{-3} \) H
• Resistance (\( R \)) = 1 k\(\Omega\) = \( 1 \times 10^{3} \) \(\Omega\)
• Electromotive Force (\( \epsilon \)) = 1 V
• Initial condition: \( i(0) = 0 \)

The solution to this first-order, linear differential equation is derived using integrating factor methods, but it's commonly known as:

\(i(t) = \frac{\epsilon}{R} \left( 1 - e^{-\frac{R}{L}t} \right)\)

Substitute the given values into the equation:

\(i(t) = \frac{1}{1000} \left( 1 - e^{-\frac{1000}{1 \times 10^{-3}} \cdot t} \right)\)

Simplify the exponent:

\(i(t) = 0.001 \left( 1 - e^{-1000000t} \right)\)

Substitute \( t = 1 \, \mu s = 1 \times 10^{-6} \, s \):

\(i(1 \, \mu s) = 0.001 \left( 1 - e^{-1000000 \times 1 \times 10^{-6}} \right)\)

\(i(1 \, \mu s) = 0.001 \left( 1 - e^{-1} \right)\)

Therefore, the current \( i(1 \, \mu s) \) in mA is:

\(i(1 \, \mu s) = (1 - e^{-1}) \, \text{mA}\)

The correct answer is 1-e^-1, which matches the provided correct answer.