Q107 / 150 · MHT CET 2023 (09 May Shift 2) · Physics

An e.m.f. $E=4 \cos (1000 t)$ volt is applied to an LR circuit of inductance $3 \mathrm{mH}$ and resistance $4 \Omega$. The maximum current in the circuit is

  • A
    $\frac{4}{\sqrt{7}} \mathrm{~A}$
  • B
    $1.0 \mathrm{~A}$
  • C
    $\frac{4}{7} \mathrm{~A}$
  • Correct answer
    D
    $0.8 \mathrm{~A}$

Solution

For LR circuit, $Z=\sqrt{R^2+(\omega L)^2}$ Comparing equation $E=4 \cos (1000 t)$ with standard equation, $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$, we get $\omega=1000$ units and $E_0=4 \mathrm{~V}$ $\begin{aligned} \therefore \quad Z & =\sqrt{16+\left(1000 \times 3^2 \times 10^{-6}\right)} \\ & Z=\sqrt{16+9}=5 \Omega \\ \therefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 \mathrm{~A} \end{aligned}$

Hence, the correct answer is option: 'D'.

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