In the circuit shown in the figure, the AC source gives a voltage...

In the circuit shown in the figure, the

\mathrm{AC}

source gives a voltage

V=20 \cos (2000 t)

neglecting source resistance, the voltmeter and ammeter reading will be In the circuit shown in the figure, the AC source gives a voltage V=20 (2000 t) neglecting source resistance, the

In the circuit shown in the figure, the AC source gives a voltage V=20 (2000 t) neglecting source resistance, the

Options

A $0 \mathrm{ V}, 0.47 \mathrm{ A}$
B $282 \mathrm{ V}, 1.41 \mathrm{ A}$
C $1.41 \mathrm{ V}, 0.47 \mathrm{ A}$
D $15 \mathrm{ V}, 837 \mathrm{ A}$

Correct answer

B. $282 \mathrm{ V}, 1.41 \mathrm{ A}$

Step-by-step solution

Given,

R_1=8 \Omega

\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}

and

\quad V_0=20 \cos (2000 t)

Impedance,

Z=\sqrt{R^2+

\left(X_L-X_C\right)^2}

As, here,

X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega

Similarly,

X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega\because \quad X_L=X_C

, Hence,

i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{ A}

Hence,

\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{ A}

and

\begin{aligned} V & =R_2 i_{\mathrm{rms}}=1.41 \times2 \\ & =282 \mathrm{ V} \end{aligned}

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