Alternating Current MCQs for NEET — Physics Questions with Answers

Figure 7.8(a) and the accompanying description clearly state: 'Here the current phasor I is $\pi/2$ ahead of the voltage phasor V as they rotate counterclockwise.' This aligns with the mathematical derivation that current leads voltage by $\pi/2$.

According to the provided text, 'We see that the current reaches its maximum value later than the voltage by one-fourth of a period $T/4 = \pi/(2\omega)$'.

The text states, 'So, the average power over a complete cycle is $(P_L) = \text{average of } [- (v_m i_m/2) \sin(2\omega t)]$ = 0, since the average of $\sin (2\omega t)$ over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero'.

The example calculates inductive reactance as $X_L = 2\pi \nu L$. This formula clearly shows that inductive reactance is directly proportional to the frequency ($\nu$) of the AC source.

The text mentions, 'in the case of an inductor, the current lags the voltage by $\pi/2$'. This is equivalent to saying voltage leads current by $\pi/2$.

From Example 7.2: $X_L = 2\pi \nu L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} \Omega = 7.85 \Omega$.

The text states, 'In order to show phase relationship between voltage and current in an ac circuit, we use the notion of phasors. The analysis of an ac circuit is facilitated by the use of a phasor diagram.'

From the section 'AC VOLTAGE APPLIED TO AN INDUCTOR', it is stated that 'the self-induced Faraday emf in the inductor' is $L \frac{di}{dt}$. Applying Kirchhoff's loop rule for a pure inductor, $v - L \frac{di}{dt} = 0$, hence $v = L \frac{di}{dt}$.

Example 7.5 states, 'As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases.'

As per Example 7.5: 'As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.'