Question 388: AC fundamentals

Question

In the circuit shown in the figure, a.c. source gives voltage $$\mathrm{V}=20 \cos (2000 \mathrm{t})$$. Impedance and r.m.s. current respectively will be

Accepted Answer

$$\begin{aligned}
& \mathrm{V}=20 \cos 2000 \mathrm{t} \quad 	herefore \omega=2000 \mathrm{~rad} / \mathrm{s} \
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2000 	imes 5 	imes 10^{-3}=10 \Omega \
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2000 	imes 50 	imes 10^{-6}}=10 \Omega
\end{aligned}$$
$$	herefore \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$$ which gives resonance.
$$\mathrm{Z}=\mathrm{R}=10 \Omega$$
[Here we are ignoring resistance of the coil which is given as $$5 \Omega$$. Otherwise, the total resistance of the circuit will be $$15 \Omega$$]
The peak current $$\mathrm{I}_0=\frac{\mathrm{V}_0}{\mathrm{R}}=\frac{20}{10}=2 \mathrm{~A}$$
R.M. S. current $$I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{~A}$$

Answer

$$10 \Omega, 0.5 \mathrm{~A}$$

Answer

$$5 \Omega, 2 \mathrm{~A}$$

Answer

$$10 \Omega, \sqrt{2} \mathrm{~A}$$

Answer

$$5 \Omega, 1 \mathrm{~A}$$

AC fundamentals

In the circuit shown in the figure, a.c. source gives voltage $\mathrm{V}=20 \cos (2000 \mathrm{t})$. Impedance and r.m.s. current respectively will be

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