Alternating Current MCQs for NEET — Physics Questions with Answers

From the NCERT text, 'In terms of I, the average power, denoted by P is \(P = I^2 R\)' (Eq. 7.7). It also states, 'Equation (7.7) can also be written as \(P = V^2 / R = I V\) (since \(V = I R\)'). Here, I and V represent rms values. Therefore, all options represent valid formulas for average power in a purely resistive AC circuit.

The 'Points to Ponder' section explicitly states, '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.' In the example, 220 V is also used as an RMS value.

The NCERT text states that the relation between RMS voltage (V) and peak voltage (\(V_m\)) is \(V = V_m / \sqrt{2}\). Therefore, \(V_m = V \sqrt{2}\). Given \(V = 220 V\), then \(V_m = 220 \times 1.414 \approx 311 V\).

Equation 7.6 defines RMS current: '\(I = i_m / \sqrt{2} = 0.707 i_m\)'.

The NCERT states, 'This shows the advantage of introducing the concept of rms values. 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.'

The NCERT states, 'Joule heating is given by \(i^2R\) and depends on \(i^2\) (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor.'

The NCERT states, 'The magnitudes of phasors V and I represent the amplitudes or the peak values \(v_m\) and \(i_m\) of these oscillating quantities.' The direction of the phasor indicates the phase.

Given \(P = 100 W\) and \(V_{rms} = 220 V\). We know \(P = V_{rms}^2 / R\). Therefore, \(R = V_{rms}^2 / P = (220 V)^2 / 100 W = 48400 / 100 = 484 \Omega\). This is directly from Example 7.1 (a).

Given \(P = 100 W\) and \(V_{rms} = 220 V\). We know \(P = I_{rms} V_{rms}\). Therefore, \(I_{rms} = P / V_{rms} = 100 W / 220 V \approx 0.454 A\). This is directly from Example 7.1 (c).

The NCERT text derived this: 'Using the trigonometric identity, \(sin^2 \omega t = 1/2 (1– cos 2\omega t)\), we have \(\langle sin^2 \omega t \rangle = (1/2) (1– \langle cos 2\omega t \rangle)\) and since \(\langle cos 2\omega t \rangle = 0\), we have, \(\langle sin^2 \omega t \rangle = 1/2\)' (Eq. 7.5c).