Animated Solution for Physics - Alternating Current: The phase difference between the alternating current and emf is $\frac{\pi }{2}$. Which of the following cannot be the constituent of the circuit ?

Have you ever wondered what it truly means when we say the phase difference between voltage and current is exactly $\frac{\pi }{2}$? It sounds like a dry mathematical fact, but physically, it represents a perfect, lossless dance of energy. Let's dive deep into this beautiful concept and uncover why certain circuit elements can perform this dance, while others simply cannot.

When an alternating current flows through a circuit, the voltage and current are not always in sync. They can peak at different times. The time delay between their peaks is represented by the phase angle, $\varphi$.

When the problem states that the phase difference is exactly $\frac{\pi }{2}$ (or $90^{\circ }$), it is giving us a massive clue. A phase difference of $\frac{\pi }{2}$ means the power factor, $\cos \varphi$, is exactly zero.

This implies that the circuit consumes absolutely zero average power! The energy is merely borrowed from the source to build an electric or magnetic field, and then returned completely. For this to happen, the circuit must be purely reactive. There can be no resistance ($R=0$) because a resistor would dissipate energy as heat, making the power factor non-zero.

Let's evaluate each option using the powerful tool of phasor diagrams.

1. The Pure Capacitor (C alone) Imagine a circuit with only a capacitor. The current leads the voltage by exactly $90^{\circ }$. If we draw the current phasor along the positive x-axis, the voltage phasor points straight down along the negative y-axis. The phase difference is exactly $\frac{\pi }{2}$. So, a pure capacitor fits the description perfectly.

2. The Pure Inductor (L alone) Now, consider a pure inductor. Here, the voltage leads the current by exactly $90^{\circ }$. The voltage phasor points straight up along the positive y-axis. Once again, the magnitude of the phase difference is exactly $\frac{\pi }{2}$. This is also a valid constituent.

3. The L-C Circuit What happens if we combine an inductor and a capacitor, but still have no resistor? The inductive voltage points up, and the capacitive voltage points down. They are exactly $180^{\circ }$ out of phase with each other. The net voltage is simply their difference, $V_L-V_C$.

Because there is no horizontal (resistive) component, the net voltage phasor will still lie entirely on the y-axis (either pointing up or down, depending on which reactance is larger). Therefore, the phase difference remains exactly $\frac{\pi }{2}$. An L-C circuit is a perfectly valid answer!

The R-L Circuit

Finally, we arrive at the R-L circuit. This circuit contains both a resistor and an inductor.

When we draw the phasor diagram, the resistive voltage $V_R$ lies along the x-axis (in phase with the current), and the inductive voltage $V_L$ lies along the y-axis. The net voltage $V_{net}$ is the vector sum of these two perpendicular components.

Using simple trigonometry, the phase angle $\varphi$ is given by:

For the phase angle to be exactly $\frac{\pi }{2}$, $\tan \varphi$ must be infinity. This can only happen if the denominator, $R$, is exactly zero. But in an R-L circuit, $R$ is non-zero!

Because $R>0$, the net voltage phasor is pulled away from the y-axis. It forms the hypotenuse of a right-angled triangle, and the angle $\varphi$ is strictly less than $90^{\circ }$.

The question asks which of the following cannot be the constituent of the circuit. Based on our rigorous phasor analysis, the R-L circuit is the only one that fails the $\frac{\pi }{2}$ test.

This problem is a fantastic reminder of the power of phasor diagrams. By simply visualizing the vectors, complex AC circuit problems become intuitive geometric puzzles. Keep practicing, and soon you'll be seeing phasors everywhere!