In the given circuit the AC source has ω = 100 rad s⁻¹. Considering the inductor and capacitor to be

Q: Why can't we just add the currents $I_1$ and $I_2$ directly to get the total current?

In AC circuits, currents and voltages are not just magnitudes; they have phase angles. Because the upper branch is capacitive and the lower is inductive, their currents peak at different times. You must add them as vectors (phasors) to account for this phase difference.

Q: How do we know the phase angle is exactly $45^\circ$ for both branches?

The phase angle $\phi$ is determined by $\tan\phi = \frac{X}{R}$. In the upper branch, $X_C = 100\ \Omega$ and $R_1 = 100\ \Omega$, so $\tan\phi_1 = 1$, which means $\phi_1 = 45^\circ$. The same logic applies to the lower branch where $X_L = 50\ \Omega$ and $R_2 = 50\ \Omega$.

Q: Why does $I_1$ lead and $I_2$ lag the voltage?

This is a fundamental property of the components. In a capacitor, current leads voltage because charge must flow before voltage can build up. In an inductor, current lags voltage because the induced EMF opposes the change in current.

Q: What if the phase difference between the two currents wasn't exactly $90^\circ$?

If the angle between the phasors was not $90^\circ$, you couldn't use the simple Pythagorean theorem. You would have to use the law of cosines: $I = \sqrt{I_1^2 + I_2^2 + 2I_1I_2\cos\theta}$, where $\theta$ is the angle between them.

Q: Why is the correct answer not in the options?

Sometimes exam questions contain errors or typos in their options. The option $4.24\text{ A}$ is a 'trap' answer designed to catch students who incorrectly use scalar addition. The mathematically rigorous answer is $3.16\text{ A}$.

Question

In the given circuit the AC source has $\omega = 100 	ext{ rad s}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?

Options:

(A) 5.9 A

(B) 4.24 A

(C) 0.94 A

(D) 6 A

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Let's embark on a journey through this fascinating parallel AC circuit. We are presented with a voltage source of $200	ext{ V}$ operating at an angular frequency $\omega = 100	ext{ rad/s}$. This source feeds into two distinct parallel branches.

The upper branch is an RC circuit, containing a capacitor $C = 100\ \mu	ext{F}$ and a resistor $R_1 = 100\ \Omega$. The lower branch is an RL circuit, featuring an inductor $L = 0.5	ext{ H}$ and a resistor $R_2 = 50\ \Omega$. Our ultimate mission is to determine the total current $I$ drawn from the source.

### The Upper Branch: Capacitive Reactance
To find the current in any branch, we must first determine its total opposition to the AC flow, known as impedance ($Z$). For the upper branch, we start by calculating the capacitive reactance ($X_C$):

$$X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 100 	imes 10^{-6}} = 100\ \Omega$$

Now, we combine this with the resistance to find the impedance $Z_1$:

$$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega$$

The current $I_1$ flowing through this branch is simply the voltage divided by the impedance:

$$I_1 = \frac{V}{Z_1} = \frac{200}{100\sqrt{2}} = \sqrt{2}	ext{ A}$$

Because this is an RC circuit, the current leads the voltage. The phase angle $\phi_1$ is given by $\cos\phi_1 = \frac{R_1}{Z_1} = \frac{1}{\sqrt{2}}$, which means $\phi_1 = 45^\circ$.

### The Lower Branch: Inductive Reactance
Now, let's shift our focus to the lower RL branch. First, we calculate the inductive reactance ($X_L$):

$$X_L = \omega L = 100 	imes 0.5 = 50\ \Omega$$

The impedance $Z_2$ for this branch is:

$$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega$$

The current $I_2$ in the lower branch is:

$$I_2 = \frac{V}{Z_2} = \frac{200}{50\sqrt{2}} = 2\sqrt{2}	ext{ A}$$

In an RL circuit, the current lags the voltage. The phase angle $\phi_2$ is $\cos\phi_2 = \frac{R_2}{Z_2} = \frac{1}{\sqrt{2}}$, meaning $\phi_2 = 45^\circ$ lagging.

### The Master Equation: Phasor Addition
Here is where many students fall into a classic trap. You might be tempted to simply add the two currents together: $I_1 + I_2 = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2} \approx 4.24	ext{ A}$. **Do not do this!** In AC circuits, currents are phasors (rotating vectors) and must be added vectorially.

If we look at our phasor diagram, $I_1$ is $45^\circ$ above the voltage reference line, and $I_2$ is $45^\circ$ below it. The total angle between them is exactly $90^\circ$. Because they are perpendicular, we can use the Pythagorean theorem to find the magnitude of the total current $I$:

$$I = \sqrt{I_1^2 + I_2^2} = \sqrt{(\sqrt{2})^2 + (2\sqrt{2})^2}$$

$$I = \sqrt{2 + 8} = \sqrt{10} \approx 3.16	ext{ A}$$

### Final Conclusion
The actual current flowing through the circuit is $3.16	ext{ A}$. If we look at the given options, none of them match this value. The option $4.24	ext{ A}$ is a deliberate distractor for those who incorrectly use scalar addition. Therefore, this question has no correct option provided.

Animated Solution for Physics - Alternating Current: In the given circuit the AC source has $ω = 100 rad s^{- 1}$ . Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?

Select Answer:

Visualized Solution

Analyzing the Setup

Impedance Formulas

Upper Branch Impedance

Upper Branch Current & Phase

Lower Branch Impedance

Lower Branch Current & Phase

Total Current Calculation

The Way Forward

The Sigma Insight: AC Circuits and Power in AC Circuits

Analyzing the Setup Let's embark on a journey through this fascinating parallel AC circuit

The Upper Branch

The Lower Branch

The Master Equation

Final Conclusion The actual current flowing through the circuit is $3.16 A$

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