Question

Find the value for $i(t)$ from the circuit.

• A. 20cos(300t + 68.2$^\circ$) A
• B. 20cos(300t - 68.2$^\circ$) A
• C. 2.48cos(300t + 68.2$^\circ$) A
• D. 2.48cos(300t - 68.2$^\circ$) A

Hint:

Always convert sinusoidal sources and reactive components into phasor and impedance forms before solving AC circuits.

Answer 1

The Correct Option is D

Step 1: Given sinusoidal voltage source
\( V(t) = 20 \cos(300t) \Rightarrow V = 20\angle0^\circ \) V (phasor form)

Step 2: Impedance of inductor
\( Z_L = j\omega L = j(300)(25 \times 10^{-3}) = j7.5\ \Omega \)

Step 3: Total impedance
\( Z = R + j\omega L = 3 + j7.5 \)
\( |Z| = \sqrt{3^2 + 7.5^2} = \sqrt{9 + 56.25} = \sqrt{65.25} \approx 8.08\ \Omega \)
\( \angle Z = \tan^{-1} \left( \frac{7.5}{3} \right) = 68.2^\circ \)

Step 4: Current (phasor)
\( I = \frac{V}{Z} = \frac{20\angle0^\circ}{8.08\angle68.2^\circ} = 2.48\angle -68.2^\circ \)

Step 5: Convert back to time domain
\( i(t) = 2.48 \cos(300t - 68.2^\circ) \)

Conclusion: The current is \( i(t) = 2.48 \cos(300t - 68.2^\circ) \) A