Correct Answer is (2)

X_L = 2$\pi$fL

f is very large

$\therefore$ X_L is very large hence open circuit.

$X_C=\frac{1}{2\pi fC}$

f is very large.

$\therefore$ X_C is very small, hence short circuit.

Final circuit



$\therefore$ $Z_{eq}=1+\frac{2\times 2}{2+2}=2$

Correct Answer is (1)

$\cos \varphi =\frac{R}{\sqrt{R^2+3R^2}}$

$=\frac{1}{\sqrt{10}}$

$\cos {\varphi }^{\prime }=\frac{R}{\sqrt{R^2+R^2}}$

$=\frac{1}{\sqrt{2}}$

$\frac{\cos {\varphi }^{\prime }}{\cos \varphi }=\frac{\sqrt{10}}{\sqrt{2}}=\frac{\sqrt{5}}{1}$

$\therefore$ x = 1

Correct Answer is (11)

$f_{rms}^2=f_{1rms}^2+f_{2rms}^2$

$={\left(\frac{\sqrt{42}}{\sqrt{2}}\right)}^2+{10}^2$

$=121\Rightarrow f_{rms}$ = 11 A

Correct Answer is (3)

Given,

Inductance, L = 30 mH

Resistance, R = 1 $\mathrm{\Omega }$

Angular frequency, $\omega$ = 300 rad/s

We know that in L-C-R circuit, $\tan \varphi =\frac{X_C-X_L}{R}$

where, $\varphi$ = phase angle = 45${}^{\circ }$

X_C = capacitive reactance = $\frac{1}{\omega C}$

X_L = inductive reactance = $\omega$L

$\Rightarrow$ $\tan {45}^{\circ }=\frac{X_C-X_L}{R}$

$\Rightarrow X_C-X_L=R$ [$\because$ tan 45${}^{\circ }$ = 1]

$\Rightarrow \frac{1}{\omega C}-\omega L=R\Rightarrow \frac{1}{\omega C}-300\times 30\times {10}^{-3}=1$

$\Rightarrow \frac{1}{\omega C}=10\Rightarrow \omega C=\frac{1}{10}$

$\Rightarrow C=\frac{1}{10\omega }\Rightarrow C=\frac{1}{10\times 300}$

$\Rightarrow C=\frac{1}{3}\times {10}^{-3}F$ .... (i)

According to question, the value of capacitance is $\frac{1}{x}\times {10}^{-3}F$. So, on comparing it with Eq. (i), we can say x = 3.

Correct Answer is (4)

At resonance power (P)

$P=\frac{{(V_{rms})}^2}{R}$

$\therefore$ $P=\frac{{(250/\sqrt{2})}^2}{8}$

$\Rightarrow$ P = 3906.25 w

$\Rightarrow$ P $\cong$ 4 Kw