Question 386: AC fundamentals

Question

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

Accepted Answer

For LR circuit,
$$\mathrm{Z}=\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}$$
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 $$E_0=4 \mathrm{~V}$$
$$\begin{aligned}
	herefore \quad & Z=\sqrt{16+\left(1000 	imes 3^2 	imes 10^{-6}ight)} \
& Z=\sqrt{16+9}=5 \Omega \
	herefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 A
\end{aligned}$$

Answer

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

Answer

$$1.0 \mathrm{~A}$$

Answer

$$\frac{4}{7} \mathrm{~A}$$

Answer

$$0.8 \mathrm{~A}$$

AC fundamentals

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

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