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

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$\frac{4}{\sqrt{7}} A$

$1.0 A$

$\frac{4}{7} A$

$0.8 A$

answer is D.

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Detailed Solution

$i = \frac{E}{z}, z = \sqrt{X_{L}^{2} + R^{2}}$

$\begin{array}{l} Z = \sqrt{R^{2} + ω^{2} L^{2}} \\ = \sqrt{(4)^{2} + {(1000 \times 3 \times 10^{- 3})}^{2}} = 5 Ω \end{array}$

$I = \frac{V}{Z} = \frac{4}{5} = 0.8 A$

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