A circuit connected to an AC source of emf e = e_0 sin(100t) with t in seconds, gives a phase differ

Q: Why does a phase difference of $\pi/4$ mean $R = X$?

In an impedance triangle, the tangent of the phase angle $\phi$ is the ratio of reactance $X$ to resistance $R$. Since $\tan(\pi/4) = 1$, it directly implies $X = R$.

Q: How do we know if it's an RC or RL circuit just from the phase difference?

The problem states 'a phase difference of $\pi/4$' without specifying if the current leads or lags the emf. Therefore, it could be either an RC circuit (current leads) or an RL circuit (current lags). We must test both possibilities.

Q: What is the formula for capacitive and inductive reactance?

Capacitive reactance is $X_C = \frac{1}{\omega C}$, and inductive reactance is $X_L = \omega L$, where $\omega$ is the angular frequency.

Q: Can we solve this by finding the impedance $Z$ first?

While you could calculate $Z = \sqrt{R^2 + X^2}$, it's an unnecessary extra step. The phase angle gives a direct relationship between $R$ and $X$, which is much faster to use for checking the options.

Question

A circuit connected to an AC source of emf $e = e_0 \sin(100t)$ with $t$ in seconds, gives a phase difference of $\frac{\pi}{4}$ between the emf $e$ and current $i$. Which of the following circuits will exhibit this?

Options:

(A) RC circuit with $R = 1	ext{ k}\Omega$ and $C = 1\mu	ext{F}$

(B) RL circuit with $R = 1	ext{ k}\Omega$ and $L = 1	ext{ mH}$

(C) RC circuit with $R = 1	ext{ k}\Omega$ and $C = 10\mu	ext{F}$

(D) RL circuit with $R = 1	ext{ k}\Omega$ and $L = 10	ext{ mH}$

Sigma Percentile · Accepted Answer

### The Dance of Voltage and Current

In the fascinating world of Alternating Current (AC), voltage and current are like two dancers. Sometimes they move perfectly in sync, and other times, one leads while the other follows. This lead or lag is what we call the **phase difference**.

In our problem, we are given an AC source with an electromotive force described by the equation $e = e_0 \sin(100t)$. We are also told that the phase difference between the emf and the current is exactly $\frac{\pi}{4}$ radians (or $45^\circ$). Our mission is to play detective and figure out which combination of circuit components creates this specific phase shift.

### The Impedance Triangle: Our Visual Guide

To understand phase difference, we turn to our trusty tool: the **impedance triangle**. In this right-angled triangle, the base represents the resistance $R$, the perpendicular height represents the net reactance $X$ (which could be inductive $X_L$ or capacitive $X_C$), and the hypotenuse represents the total impedance $Z$.

The relationship between these quantities is beautifully captured by trigonometry:
$$ 	an\phi = \frac{X}{R} $$

Since we are given that $\phi = \frac{\pi}{4}$, we can substitute this into our equation:
$$ 	an\left(\frac{\pi}{4}ight) = 1 $$

This leads us to a crucial revelation:
$$ \frac{X}{R} = 1 \implies X = R $$

This means that for our circuit to have a phase difference of $\frac{\pi}{4}$, the net reactance must be exactly equal to the resistance!

### Decoding the EMF Equation

Now, let's extract more clues from the given emf equation, $e = e_0 \sin(100t)$. The standard form of an alternating emf is $e = e_0 \sin(\omega t)$, where $\omega$ is the angular frequency.

By simply comparing the two equations, we can immediately see that:
$$ \omega = 100 	ext{ rad/s} $$

This angular frequency is the heartbeat of our circuit, and it will help us test the given options.

### The Ultimate Test: Checking the Options

We know that $X = R$ and $\omega = 100 	ext{ rad/s}$. Let's formulate our test conditions.

If the circuit is an RC circuit, the capacitive reactance is $X_C = \frac{1}{\omega C}$. Setting this equal to $R$ gives us:
$$ \omega = \frac{1}{RC} $$

If the circuit is an RL circuit, the inductive reactance is $X_L = \omega L$. Setting this equal to $R$ gives us:
$$ \omega = \frac{R}{L} $$

Now, we just need to plug the values from the options into these formulas and see which one yields $\omega = 100 	ext{ rad/s}$.

Let's test **Option (c)**, which proposes an RC circuit with $R = 1 	ext{ k}\Omega$ and $C = 10 \mu	ext{F}$.
First, let's convert these to standard units:
$R = 10^3 \Omega$
$C = 10 	imes 10^{-6} 	ext{ F} = 10^{-5} 	ext{ F}$

Now, let's calculate the angular frequency:
$$ \omega = \frac{1}{RC} = \frac{1}{10^3 	imes 10^{-5}} $$
$$ \omega = \frac{1}{10^{-2}} = 100 	ext{ rad/s} $$

Bingo! The calculated angular frequency perfectly matches the frequency from our emf equation. Therefore, this specific RC circuit is the correct answer.

### The Beauty of Special Cases

Problems like this highlight the elegance of AC circuit analysis. By understanding the geometric relationship between resistance and reactance, we can quickly deduce the behavior of the circuit.

Consider a different scenario: what if the phase difference was $0$? This would mean $	an(0) = 0$, so the net reactance $X$ must be zero. This occurs when the inductive and capacitive reactances perfectly cancel each other out ($X_L = X_C$), a phenomenon known as **resonance**. Always keep these special cases in your toolkit!

Animated Solution for Physics - Alternating Current: A circuit connected to an AC source of emf $e = e_{0} sin (100 t)$ with $t$ in seconds, gives a phase difference of $\frac{π}{4}$ between the emf $e$ and current $i$ . Which of the following circuits will exhibit this?

Select Answer:

Visualized Solution

Analyzing the Given Data

Phase Difference Formula

Condition for $ϕ = \frac{π}{4}$

Extracting Angular Frequency

Formulating the Check

Testing Option (c)

Conclusion

The Way Forward

The Sigma Insight: AC Circuits and Power in AC Circuits

The Dance of Voltage and Current

The Impedance Triangle

Decoding the EMF Equation

The Ultimate Test

The Beauty of Special Cases

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