Question

In the circuit shown in the figure, the $\text{AC}$ source gives a voltage $V=20\cos 2000t$. If the reading of voltmeter is $V_1$, ammeter is $A_1$ and the power consumed by the circuit is $P$, then:

• A. $V_1=\sqrt{2}\text{V}$
• B. $A_1=\sqrt{2}\text{A}$
• C. $P=20\text{W}$
• D. $A_1=4\sqrt{2}\text{V}$

Answer 1

Answer: (B), (C)

$P=20\text{W}$


Given:
$V=20\cos \left(2000t\right)$

$\Rightarrow V_0=20\text{V},\omega =2000\text{rad/sec}$

The net resistance of the circuit is:

$R=6+4=10\Omega$

The reactance of capacitor and inductor are given by:

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

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

$\therefore$ The impedance of the circuit is:

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{{10}^2+(10-10{)}^2}$

$\Rightarrow Z=10\Omega$

The maximum current flowing in the circuit is:

$i_0=\frac{V_0}{Z}=\frac{20}{10}=2\text{A}$

Ammeter reads $\text{rms}$ value of current. Hence, reading of ammeter is given by:

$A_1=i_{rms}=\frac{i_0}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\text{A}$

$[\because \text{Impedance across voltmeter is}{Z}_1=\sqrt{4^2+(10-10{)}^2}]$

$\therefore$ The reading of voltmeter is:

$V_1=i_{rms}\times \sqrt{4^2+(10-10{)}^2}$

$=\sqrt{2}\times 4=4\sqrt{2}\text{V}$

The power consumed by the circuit is:

$P=V_{rms}{i}_{rms}\cos \varphi$

$=\frac{20}{\sqrt{2}}\times \sqrt{2}\times \frac{10}{10}[\because \cos \varphi =\frac{R}{Z}]$

$=20\text{W}$

Hence, options $\left(B\right)$ and $\left(C\right)$ are the correct answers.