(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 $L-C-R$ series circuit is connected to an $AC$ source of peak voltage $240 \ V$. The phase difference between voltage and current of this circuit is $45^{\circ}$ and the resistance is $100 \ \Omega$. The $rms$ value of current through the circuit is . . . . . . . (in $A$)

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In an $LCR$ circuit,the potential difference between the terminals of the inductor is $60\,V$,between the terminals of the capacitor is $30\,V$,and between the terminals of the resistor is $40\,V$. The supply voltage will be equal to ......$V$.

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In a series $LCR$ circuit,$R$ represents the resistance of an electric bulb. If the frequency of the $A.C.$ supply is doubled,what should be the values of inductance $L$ and capacitance $C$ to maintain the same current?

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As shown in the figure,a series circuit connected across a $200 \,V, 60 \,Hz$ line consists of a capacitor of capacitive reactance $X_C = 30 \,\Omega$,a non-inductive resistor of $R_1 = 44 \,\Omega$,and a coil of inductive reactance $X_L = 90 \,\Omega$ and resistance $R_2 = 36 \,\Omega$. The power dissipated in the coil is.....$W$

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In an $AC$ circuit,an inductor,a capacitor,and a resistor are connected in series with $X_{L} = R = X_{C}$. The impedance of this circuit is:

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An $L-C-R$ series circuit is connected to an $AC$ source of peak voltage $240 \ V$. The phase difference between voltage and current of this circuit is $45^{\circ}$ and the resistance is $100 \ \Omega$. The $rms$ value of current through the circuit is . . . . . . . (in $A$)

In an $LCR$ circuit,the potential difference between the terminals of the inductor is $60\,V$,between the terminals of the capacitor is $30\,V$,and between the terminals of the resistor is $40\,V$. The supply voltage will be equal to ......$V$.

In a series $LCR$ circuit,$R$ represents the resistance of an electric bulb. If the frequency of the $A.C.$ supply is doubled,what should be the values of inductance $L$ and capacitance $C$ to maintain the same current?

In an $AC$ circuit,an inductor,a capacitor,and a resistor are connected in series with $X_{L} = R = X_{C}$. The impedance of this circuit is:

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