Alternating Current MCQs for NEET — Physics Questions with Answers

The text states: 'In terms of I, the average power, denoted by P is $P = I_m^2 R/2 = I^2 R$'. It also mentions, 'In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case.' This supports $P=I^2R$ as the equivalent expression.

As stated in 'POINTS TO PONDER 1.', 'When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage.' Similarly, the 220V rating for the bulb in example 7.1 is an RMS value.

From Example 7.1(a): 'We are given P = 100 W and V = 220 V. The resistance of the bulb is $R = V^2 / P = (220 V)^2 / 100 W = 48400 / 100 \Omega = 484 \Omega$'.

From Example 7.1(c): 'Since, $P = I V$, then $I = P / V = 100 W / 220 V = 0.454 A$'.

The context states: 'Thus, the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero.' This is because for every positive value, there is a corresponding negative value.

The table for physical quantities and remarks states: 'rms voltage $V = \frac{V_m}{\sqrt{2}}$, $V_m$ is the amplitude of the ac voltage.' This directly provides the relationship.

The 'POINTS TO PONDER 4.' section states: 'Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions.'

From 'POINTS TO PONDER 2.', 'The power rating of an element used in ac circuits refers to its average power rating.' From 'POINTS TO PONDER 3.', 'The power consumed in an ac circuit is never negative.' And from the example 7.2 explanation, 'Thus, the average power supplied to an inductor over one complete cycle is zero.' However, Case (iv) under power factor discusses LCR series circuit at resonance: 'At resonance Xc – XL= 0, and φ = 0. Therefore, cos φ = 1 and P = I^2Z = I^2R. That is, maximum power is dissipated in a circuit (through R) at resonance.' Thus, option 4 is incorrect.

The context states: 'The instantaneous power dissipated in the resistor is $p = i^2 R = i_m^2 R \sin^2 (\omega t)$'. This directly provides the formula for instantaneous power.

According to the NCERT text, 'In a RLC circuit, resonance phenomenon occur when $X_L = X_C$ or $\omega_0 = 1/\sqrt{LC}$. For resonance to occur, the presence of both L and C elements in the circuit is a must.'