Question

In a L-R circuit of 4 mH inductance and \(3 \Omega\) resistance, e.m.f. \(E = cos(1000t)\) V is applied. The amplitude of current is

• A. \(1.6\) A
• B. \(1\) A
• C. \(0.2\) A
• D. \(0.4\) A

Hint:

The impedance of an LR circuit is the root of R squared plus X_L squared.

Answer 1

The Correct Option is C

Step 1: Approach
Use the 3-4-5 triangle.

Step 2: Triangle
Resistance 3 ohm and reactance $1000\times0.004=4$ ohm give the hypotenuse 5 ohm.

Step 3: Current
$I_0=\dfrac{1}{5}=0.2$ A. Option (C).

Final Answer:
The reactance is 4 ohm and the impedance is 5 ohm, so the current amplitude is 0.2 A, option (C). \[ \boxed{0.2\ \text{A}} \]