In the circuit shown,the power factor of the | vedclass

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(D) Given: $C = 1 	ext{ mF} = 10^{-3} 	ext{ F}$,$V = 90 	ext{ V}$,$\omega = 25 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C}

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$3$

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(D) Given: $C = 1 	ext{ mF} = 10^{-3} 	ext{ F}$,$V = 90 	ext{ V}$,$\omega = 25 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{25 	imes 10^{-3}} = \frac{1000}{25} = 40 \Omega$.The power factor of the circuit is $1$,which means the circuit is in resonance. Thus,the total reactance of the circuit must be zero.Since the circuit contains a capacitor,the box must contain an inductor to cancel the capacitive reactance.Let the box have resistance $R$ and inductive reactance $X_L$. For resonance,$X_L = X_C = 40 \Omega$.The power factor of the box is $\cos \phi = \frac{R}{Z_{	ext{box}}} = \frac{3}{5}$.Since $Z_{	ext{box}} = \sqrt{R^2 + X_L^2}$,we have $\frac{R}{\sqrt{R^2 + 40^2}} = \frac{3}{5}$.Squaring both sides: $\frac{R^2}{R^2 + 1600} = \frac{9}{25} \implies 25R^2 = 9R^2 + 14400 \implies 16R^2 = 14400 \implies R^2 = 900 \implies R = 30 \Omega$.The total impedance of the circuit is $Z = R = 30 \Omega$ (since $X_L - X_C = 0$).The current in the circuit is $I = \frac{V}{Z} = \frac{90}{30} = 3 	ext{ A}$.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

In the circuit shown,the power factor of the circuit is $1$ and the power factor of the box is $\frac{3}{5}$. Find the reading of the ammeter in $A$.

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