NEET Alternating Current MCQs & PYQs

Practice NEET Alternating Current Questions

Question 1:

easy

In pure inductive circuit, the curves between frequency f and reciprocal of inductive reactance 1/XL is :

Question 2:

easy

In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E = E0 cos (wt) applied to the circuit. The power consumed in the circuit is :

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Question 3:

easy

Which one of the following curves represents the variation of impedence (Z) with frequency f in series LCR circuit :-

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Question 4:

easy

In series LCR circuit voltage across L, C & R is 20V each. If capacitor is short-circuited then
voltage across inductor is :

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Question 5:

easy

A coil has power factor of 0.707 at 60 Hz. Then its power factor at 180 Hz will be :-

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Question 6:

easy

A variable inductor is connected to an ac source. What effect does increasing the inductance have on the reactance and current in this circuit ?

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Question 7:

easy

A variable capacitor is connected to an ac source. What effect does decreasing the capacitance have on the reactance and current in this circuit ?

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Question 8:

easy

Given below are two statements :
Statement I : In an LCR series circuit, current is maximum at resonance.
Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit (using same resistance) when connected to same voltage source.
In the light of the above statements, choose the correct from the options given below :

At resonance, impedance of an LCR circuit is minimum and equal to \(R\), so current is maximum. In general, impedance \(Z ge R\), hence the current \(I = V/Z\) is less than or equal to the purely resistive current \(V/R\). Thus, both statements are true.

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Question 9:

easy

The maximum power is dissipated for an ac in a/an

The average power dissipated is \(P = V_{\text{rms}} I_{\text{rms}} cos \phi\). For a purely resistive circuit, the phase angle \(\phi = 0\), which gives the maximum power factor \(cos \phi = 1\).

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Question 10:

easy

The peak value of the alternating current given by \(I = 4 \sin\omega t + 4 \sin(\omega t + 2\pi/3)\) is:

The two currents can be vectorially added with phase difference \(\phi = 2\pi/3 = 120^\circ\). Resultant amplitude is \(I_0 = \sqrt{4^2 + 4^2 + 2 \times 4 \times 4 \cos 120^\circ} = 4\).

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