Question

An inductor of inductance $2\, H$ and a resistance of $10\, \Omega$ are connected in series to an ac source of $110\, V,\, 60\, Hz$. The current in the circuit will be

• A. 0.32 A
• B. 0.15 A
• C. 0.48 A
• D. 0.80 A

Answer 1

Answer: (B)

$X_{L}=\omega L=2 \pi \cup L=2 \pi \times 60 \times 2=240 \pi \Omega,\, R=10 \Omega$
$Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{10^{2}+(240 \pi)^{2}} \approx 240 \pi$
$I_{ rms }=\frac{V_{ rms }}{Z}=\frac{110}{240 \pi}=0.15\, A$