When an AC source of emf e = E_0 sin(100t) is connected across a circuit, the phase difference betwe

Q: Why did we choose an R-C circuit and not an L-C circuit?

In an $L-C$ circuit, the phase difference between voltage and current is always exactly $\pi/2$ (or $90^\circ$). Since the given phase difference is $\pi/4$ ($45^\circ$), the circuit must contain a resistor to shift the phase angle to a value between $0$ and $\pi/2$.

Q: How do we know from the graph that current leads the voltage?

Look at the time axis. The current wave (dashed line) crosses the zero mark and reaches its positive peak at an earlier time than the voltage wave (solid line). Reaching the peak earlier means it is 'ahead' or leading in phase.

Q: What is the significance of the product CR?

The product $CR$ is known as the time constant of the circuit. In the context of AC circuits, it determines the phase shift for a given frequency. A specific phase shift requires a specific time constant.

Q: Could it be an R-L-C circuit?

The problem states the circuit consists 'possibly only of R-C or R-L or L-C'. However, even if it were an R-L-C circuit, the net reactance would have to be capacitive ($X_C > X_L$) to make the current lead, and the net reactance divided by $R$ would still need to equal 1.

Q: Why did we convert units in the final step?

Standard SI units must be used to ensure the product $CR$ matches the unitless ratio derived from $\omega$. Resistance must be in Ohms ($\Omega$) and capacitance in Farads (F) to yield a time constant in seconds.

Question

When an AC source of emf $e = E_0 \sin (100t)$ is connected across a circuit, the phase difference between the emf $e$ and the current $i$ in the circuit is observed to be $\frac{\pi}{4}$ ahead, as shown in the diagram. If the circuit consists possibly only of $R-C$ or $R-L$ or $L-C$ in series, find the relationship between the two elements.

Options:

(A) $R = 1	ext{ k}\Omega, C = 10	ext{ }\mu	ext{F}$

(B) $R = 1	ext{ k}\Omega, C = 1	ext{ }\mu	ext{F}$

(C) $R = 1	ext{ k}\Omega, L = 10	ext{ H}$

(D) $R = 1	ext{ k}\Omega, L = 1	ext{ H}$

Sigma Percentile · Accepted Answer

### Decoding the Graph

Imagine you are watching a race between two runners: Voltage and Current. In a purely resistive circuit, they run side-by-side, perfectly in sync. But the moment we introduce an inductor or a capacitor, the track changes. One runner gets a head start.

In our problem, the graph is the ultimate storyteller. Look closely at the time axis. The dashed line, representing the current $i$, reaches its positive peak **before** the solid line, which represents the electromotive force (emf) $e$. In the language of physics, we say that the **current leads the voltage**. The problem explicitly tells us that this lead is exactly $\frac{\pi}{4}$ radians (or $45^\circ$).

### The Physics of Phase Difference

Why does this happen? In an AC circuit, an inductor opposes changes in current, causing the current to lag behind the voltage. A capacitor, on the other hand, opposes changes in voltage, causing the voltage to lag behind the current (which is the same as saying the current leads the voltage).

Since our graph clearly shows the current leading, we can immediately rule out any circuit that is purely inductive or where the inductive effect dominates. The problem gives us three possibilities: $R-C$, $R-L$, or $L-C$. 
- An $R-L$ circuit would have current lagging.
- An $L-C$ circuit would have a phase difference of exactly $\frac{\pi}{2}$ or $-\frac{\pi}{2}$, depending on which component is stronger.

Because our phase difference is $\frac{\pi}{4}$ (a value strictly between $0$ and $\frac{\pi}{2}$), the circuit must be a combination of a resistor and a capacitor. We are dealing with an **$R-C$ series circuit**.

### The Mathematical Execution

Now that we have identified the circuit, let's bring in the heavy machinery. For an $R-C$ series circuit, the phase angle $\phi$ is determined by the ratio of the capacitive reactance $X_C$ to the resistance $R$:

$$ 	an \phi = \frac{X_C}{R} $$

We know that the capacitive reactance is inversely proportional to the angular frequency $\omega$ and the capacitance $C$, given by $X_C = \frac{1}{\omega C}$. Substituting this into our phase angle equation gives:

$$ 	an \phi = \frac{1}{\omega C R} $$

The problem provides us with the emf equation: $e = E_0 \sin(100t)$. Comparing this to the standard form $e = E_0 \sin(\omega t)$, we can extract the angular frequency: $\omega = 100 	ext{ rad/s}$. We also know the phase angle $\phi = \frac{\pi}{4}$. Let's plug these values into our equation:

$$ 	an\left(\frac{\pi}{4}ight) = \frac{1}{100 \cdot C \cdot R} $$

Since $	an\left(\frac{\pi}{4}ight) = 1$, the equation simplifies beautifully:

$$ 1 = \frac{1}{100 C R} $$

Rearranging this to solve for the product $CR$, we get:

$$ C R = \frac{1}{100} = 10^{-2} 	ext{ s} $$

### The Final Verification

We have found the golden rule for our circuit: the product of the resistance and the capacitance must be exactly $10^{-2}$ seconds. The final step is to test the given options to see which one obeys this rule.

Let's test **Option (a)**:
- Resistance $R = 1 	ext{ k}\Omega = 10^3 	ext{ }\Omega$
- Capacitance $C = 10 	ext{ }\mu	ext{F} = 10 	imes 10^{-6} 	ext{ F} = 10^{-5} 	ext{ F}$

Multiplying them together:
$$ R 	imes C = (10^3 	ext{ }\Omega) 	imes (10^{-5} 	ext{ F}) = 10^{-2} 	ext{ s} $$

It's a perfect match! The values in Option (a) satisfy the physical and mathematical constraints of our circuit.

Select Answer:

Visualized Solution

\text{Analyzing the Graph}

\text{Identifying the Circuit}

\text{Phase Angle Formula}

\text{Substituting Values}

\text{Calculating } CR

\text{Checking Options}

\text{Conclusion}

The Sigma Insight: Alternating Current (AC) and Voltage

Decoding the Graph

The Physics of Phase Difference

The Mathematical Execution

The Final Verification

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