A L-C-R series circuit is connected to a sour | vedclass

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(D) In an $L-C-R$ series circuit,the phase difference $\phi$ between the current and the voltage is given by the relation $\cos \phi = \frac{R}{Z}$,wh

Vedclass · Accepted Answer

$0^{\circ}$

Vedclass · Answer

(D) In an $L-C-R$ series circuit,the phase difference $\phi$ between the current and the voltage is given by the relation $\cos \phi = \frac{R}{Z}$,where $Z$ is the impedance of the circuit. At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,i.e.,$X_L = X_C$. The impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Substituting $X_L = X_C$ into the impedance formula,we get $Z = \sqrt{R^2 + 0} = R$. Now,substituting $Z = R$ into the phase difference formula,we get $\cos \phi = \frac{R}{R} = 1$. Since $\cos \phi = 1$,it follows that $\phi = 0^{\circ}$. Therefore,at resonance,the applied voltage and the current are in the same phase.

$A$ $L-C-R$ series circuit is connected to a source of alternating current. At resonance,the applied voltage and the current flowing through the circuit will have a phase difference of

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