Question

In an $LR$ circuit, the value of $L$ is $\frac{0.4}{\pi }\text{H}$ and the value $R$ is $30\Omega$. If in the circuit, an alternating $emf$ of $200\text{V}$ $rms$ value at $50\text{cycles/s}$ is connected, the peak current of the circuit is $k\sqrt{2}\text{A}$ .The value of $k$ is (integer only)

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Answer 1

Impedence for $LR$ circuit is given by,

$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+{\omega }^2{L}^2}$

As, $f=50\text{Hz}$ $\Rightarrow \omega =2\pi f=100\pi$

$\therefore X_L=\omega L=100\pi \frac{\left(0.4\right)}{\pi }=40\Omega$

So, $Z=\sqrt{{30}^2+{40}^2}=50\Omega$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{200}{50}=4\text{A}$

$\therefore I_{peak}=I_{rms}\sqrt{2}=4\sqrt{2}$

Given, $I_{peak}=k\sqrt{2}$

$\Rightarrow k=4$