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JEE PHYSICS MCQ: ALTERNATING CURRENT

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An ac ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the ac ammeter reads 3 A3 \mathrm{~A}. When another alternating current passes through the circuit, the ac ammeter reads 4 A4 \mathrm{~A}. Then find the reading of this ammeter (in A\mathrm{A}), if dc and ac flow through the circuit simultaneously.
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Correct option: A
5
Option A (5) is correct because the AC ammeter reads the RMS value of the combined DC and AC currents, which is calculated to be 5A.

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In a series RC circuit with an AC source, R=300 ΩR=300 \, \Omega, C=25 μF,ε0=50 VC=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V} and v=50/π Hzv=50 / \pi \, \mathrm{Hz}. Find the peak current and the average power dissipated in the circuit. In a series circuit R=300Ω,L=0.9 H,C=2.0R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0 μF\mu F and ω=1000 rad/sec\omega=1000 \mathrm{~rad} / \mathrm{sec}. The impedance of the circuit is: An L-C-R series circuit with a resistance of 100Ω100 \Omega is connected to 200 V200 \mathrm{~V} (AC source) and angular frequency 300 rad/s300 \, \mathrm{rad} / \mathrm{s}. When only the capacitor is removed, then the current lags behind the voltage by 60∘60^{\circ}. When only the inductor is removed the current leads the voltage by 60∘60^{\circ}. The average power dissipated in original L-C-R circuit is Assertion (A) : In series L−C⋅R,ACL-C \cdot R, \mathrm{AC} circuit, current and voltage are in same phase at resonance.

Reason (R) : In series L−C−R,ACL-C-R, \mathrm{AC} circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance.
A series L−C−RL-C-R circuit is connected to an AC\mathrm{AC} voltage source. When LL is removed from the circuit, the phase difference between current and voltage is π4\frac{\pi}{4}. If instead CC is removed from the circuit, the phase difference is again π4\frac{\pi}{4} between current and voltage. The power factor of the circuit is A capacitor in an ideal LC circuit is fully charged by a DC source. Then it is disconnected from DC source, the current in the circuit
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