An LCR series circuit with R = 100,Omega is c | vedclass

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(A) Given: $R = 100\,\Omega$,$V = 200\,V$.When capacitance is removed,the circuit becomes an $LR$ series circuit. The phase angle $\phi = 60^o$ is giv

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$2$

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(A) Given: $R = 100\,\Omega$,$V = 200\,V$.When capacitance is removed,the circuit becomes an $LR$ series circuit. The phase angle $\phi = 60^o$ is given by $	an(60^o) = \frac{X_L}{R}$.Thus,$X_L = R 	an(60^o) = 100 	imes \sqrt{3} = 100\sqrt{3}\,\Omega$.When inductance is removed,the circuit becomes an $RC$ series circuit. The phase angle $\phi = 60^o$ is given by $	an(60^o) = \frac{X_C}{R}$.Thus,$X_C = R 	an(60^o) = 100 	imes \sqrt{3} = 100\sqrt{3}\,\Omega$.In the original $LCR$ circuit,the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Since $X_L = X_C$,$Z = \sqrt{R^2 + 0} = R = 100\,\Omega$.The current in the circuit is $I = \frac{V}{Z} = \frac{200}{100} = 2\,A$.

An $LCR$ series circuit with $R = 100\,\Omega$ is connected to a $200\,V, 50\,Hz$ $a.c.$ source. When only the capacitance is removed,the current lags the voltage by $60^o$. When only the inductance is removed,the current leads the voltage by $60^o$. The current in the circuit is.....$A$

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