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(B) In the given $LCR$ series circuit,the voltage across the inductor is $V_L = 100 \ V$ and the voltage across the capacitor is $V_C = 100 \ V$.Since

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$6$

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(B) In the given $LCR$ series circuit,the voltage across the inductor is $V_L = 100 \ V$ and the voltage across the capacitor is $V_C = 100 \ V$.Since $V_L = V_C$,the circuit is in a state of resonance.In a series $LCR$ circuit at resonance,the net voltage across the inductor and capacitor is zero $(V_L - V_C = 0)$.Therefore,the entire source voltage $V$ appears across the resistor $R$.Given $V = 300 \ V$ and $R = 50 \ \Omega$.The current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R}$.Substituting the values,$I = \frac{300 \ V}{50 \ \Omega} = 6 \ A$.Thus,the reading of the ammeter is $6 \ A$.

The reading of the ammeter in the following circuit is (in $A$)

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