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(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.Given: $V_L = V_1 = 300 \, V$ and $V_C = V_2 = 300 \, V$.Since $V

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$220 \, V, \, 2.2 \, A$

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(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.Given: $V_L = V_1 = 300 \, V$ and $V_C = V_2 = 300 \, V$.Since $V_L = V_C$,the circuit is in resonance.In a series $LCR$ circuit at resonance,the net voltage across the $LC$ combination is zero $(V_{LC} = V_L - V_C = 300 - 300 = 0 \, V)$.Therefore,the entire source voltage appears across the resistor $R$.Thus,the reading of voltmeter $V_3$ is $V_R = 220 \, V$.The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.So,the reading of voltmeter $V_3$ is $220 \, V$ and the reading of ammeter $A$ is $2.2 \, A$.

In the given circuit,the readings of voltmeters $V_1$ and $V_2$ are $300 \, V$ each. The readings of the voltmeter $V_3$ and ammeter $A$ are respectively:

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