In the LCR circuit shown in the figure,what w | vedclass

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(D) In a series $LCR$ circuit,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.Given: $V = 220 \ V$,$V_L = 300 \ V$,$V_C = 300 \ V

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$220 \ V, 2.2 \ A$

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(D) In a series $LCR$ circuit,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.Given: $V = 220 \ V$,$V_L = 300 \ V$,$V_C = 300 \ V$.Substituting these values into the formula: $220 = \sqrt{V_R^2 + (300 - 300)^2}$.$220 = \sqrt{V_R^2 + 0^2} = V_R$.Thus,the reading of the voltmeter across the resistor is $V_R = 220 \ V$.The current $I$ in the circuit is given by $I = \frac{V_R}{R}$.Given $R = 100 \ \Omega$,we have $I = \frac{220 \ V}{100 \ \Omega} = 2.2 \ A$.Therefore,the readings are $220 \ V$ and $2.2 \ A$.

In the $LCR$ circuit shown in the figure,what will be the readings of the voltmeter across the resistor $(V_R)$ and the ammeter $(A)$ if an a.c. source of $220 \ V$ and $100 \ Hz$ is connected to it?

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