In an LCR circuit shown in the following figu | vedclass

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

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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 the relation:$V = \sqrt{V_R^2 + (V_L - V_C)^2}$Given from the figure:$V = 220\,V$$V_L = 300\,V$$V_C = 300\,V$Substituting these values into the equation:$220 = \sqrt{V_R^2 + (300 - 300)^2}$$220 = \sqrt{V_R^2 + 0^2}$$V_R = 220\,V$Now,the current $I$ in the circuit is given by:$I = \frac{V_R}{R}$Given $R = 100\,\Omega$:$I = \frac{220}{100} = 2.2\,A$Thus,the reading of the voltmeter is $220\,V$ and the reading of the ammeter is $2.2\,A$.

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

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