Question 189: AC fundamentals

Question

An inductor of self inductance 1 H is connected in series with a resistor of $100 \pi$ ohm and an ac supply of $100 \pi$ volt, 50 Hz . Maximum current flowing in the circuit is _________ A.

Accepted Answer

Impedance of circuit
$$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}ight)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}ight)^2} \
& =\sqrt{(100 \pi)^2+(2 \pi 	imes 50 	imes 1)^2} \
& =\sqrt{(100 \pi)^2+(100 \pi)^2} \
& =\sqrt{2} 	imes 100 \pi \
& \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}}{2}=\frac{100 \pi}{\sqrt{2} 	imes 100 \pi}=\frac{1}{\sqrt{2}} \
& \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} 	imes \frac{1}{\sqrt{2}}=1 	ext { Ampere }
\end{aligned}$$
Correct Answer : 1

AC fundamentals

An inductor of self inductance 1 H is connected in series with a resistor of 100 \pi ohm and an ac supply of 100 \pi volt, 50 Hz . Maximum current flowing in the circuit is _________ A.

Integer / Numeric Answer Type