(D) For the $L-R$ circuit: $\cos \phi_1 = \frac{R_1}{Z_1} = 0.6 = \frac{3}{5}$. Thus,$\sin \phi_1 = \frac{X_L}{Z_1} = 0.8 = \frac{4}{5}$. So,$X_L = \f

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(D) For the $L-R$ circuit: $\cos \phi_1 = \frac{R_1}{Z_1} = 0.6 = \frac{3}{5}$. Thus,$\sin \phi_1 = \frac{X_L}{Z_1} = 0.8 = \frac{4}{5}$. So,$X_L = \frac{4}{3} R_1$.For the $C-R$ circuit: $\cos \phi_2 = \frac{R_2}{Z_2} = 0.5 = \frac{1}{2}$. Thus,$\sin \phi_2 = \frac{X_C}{Z_2} = \frac{\sqrt{3}}{2}$. So,$X_C = \sqrt{3} R_2$.When joined in series,the circuit becomes an $L-C-R$ circuit. For the power factor to be $1$,the circuit must be in resonance,meaning $X_L = X_C$.Equating the reactances: $\frac{4}{3} R_1 = \sqrt{3} R_2$.Therefore,the ratio $\frac{R_1}{R_2} = \frac{3\sqrt{3}}{4}$.

Class 12 Physics Alternating Current Inductance, Capacitance and Resistance in Series and Parallel

Medium

The power factor of an $L-R$ series circuit is $0.6$ and that of a $C-R$ series circuit is $0.5$. If the elements ($L, C,$ and $R$) of the two circuits are joined in series,the power factor of this circuit is found to be $1$. The ratio of the resistance in the $L-R$ circuit to the resistance in the $C-R$ circuit is:

A
$6/5$
B
$5/6$
C
$\frac{4}{3\sqrt{3}}$
D
$\frac{3\sqrt{3}}{4}$

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Alternating Current

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Inductance, Capacitance and Resistance in Series and Parallel

An inductor of reactance $100 \text{ } \Omega$, a capacitor of reactance $50 \text{ } \Omega$, and a resistor of resistance $50 \text{ } \Omega$ are connected in series with an $AC$ source of $10 \text{ V}$, $50 \text{ Hz}$. Average power dissipated by the circuit is (in $\text{ W}$)

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In the circuit shown,$C = \frac{\sqrt{3}}{2} \times 10^{-3} \, F$,$R_2 = 20 \, \Omega$,$L = \frac{\sqrt{3}}{10} \, H$,and $R_1 = 10 \, \Omega$. The current in the $L-R_1$ branch is $I_1$ and in the $C-R_2$ branch is $I_2$. The voltage of the $A.C.$ source is given by $V = 200\sqrt{2} \sin(100t) \, V$. The phase difference between $I_1$ and $I_2$ is:

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An $AC$ source is connected in the given series $LCR$ circuit. The $RMS$ potential difference across the capacitor of $20 \mu F$ is . . . . . . $V$.
$V = 50 \sqrt{2} \sin 100 t$ volt

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$A$ circuit has a resistance of $11\,\Omega$,an inductive reactance of $25\,\Omega$,and a capacitive reactance of $18\,\Omega$. It is connected to an $AC$ source of $260\,V$ and $50\,Hz$. The current through the circuit (in amperes) is:

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In a parallel $AC$ circuit with $R_1 = R_2 = R$ and $X_L = X_C = X$. An $AC$ source $V = V_0 \sin(\omega t)$ is connected across the circuit as shown in the figure. The $RMS$ value of the potential difference $V_A - V_B$ will be:

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An inductor of reactance $100 \text{ } \Omega$, a capacitor of reactance $50 \text{ } \Omega$, and a resistor of resistance $50 \text{ } \Omega$ are connected in series with an $AC$ source of $10 \text{ V}$, $50 \text{ Hz}$. Average power dissipated by the circuit is (in $\text{ W}$)

An $AC$ source is connected in the given series $LCR$ circuit. The $RMS$ potential difference across the capacitor of $20 \mu F$ is . . . . . . $V$.$V = 50 \sqrt{2} \sin 100 t$ volt

$A$ circuit has a resistance of $11\,\Omega$,an inductive reactance of $25\,\Omega$,and a capacitive reactance of $18\,\Omega$. It is connected to an $AC$ source of $260\,V$ and $50\,Hz$. The current through the circuit (in amperes) is:

In a parallel $AC$ circuit with $R_1 = R_2 = R$ and $X_L = X_C = X$. An $AC$ source $V = V_0 \sin(\omega t)$ is connected across the circuit as shown in the figure. The $RMS$ value of the potential difference $V_A - V_B$ will be:

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An $AC$ source is connected in the given series $LCR$ circuit. The $RMS$ potential difference across the capacitor of $20 \mu F$ is . . . . . . $V$.
$V = 50 \sqrt{2} \sin 100 t$ volt