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: Understanding the Concept
The applied emf is \(E=\cos(1000t)\) V. So the peak emf is \(E_0=1\) V and \(\omega=1000\) rad/s.

Step 2: Inductive reactance
\[ X_L=\omega L=1000\times4\times10^{-3}=4\ \Omega \]

Step 3: Impedance
\[ Z=\sqrt{R^2+X_L^2}=\sqrt{3^2+4^2}=5\ \Omega \]

Step 4: Amplitude of current
\[ I_0=\frac{E_0}{Z}=\frac15=0.2\ \text{A} \]

Step 5: Check the options
The value 1 A is for a resistance of 1 ohm, and 0.4 A or 1.6 A would need an impedance of 2.5 or 0.625 ohm. The answer is 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). \[ \boxed{0.2\ \text{A}} \]