In the given circuit, the AC source has omega = 100 rad/s. Considering the inductor and capacitor to

Q: Why can't we just add the currents algebraically ($I = I_1 + I_2$)?

In AC circuits, currents and voltages have both magnitude and phase. Because the RC branch current leads the voltage and the RL branch current lags it, they peak at different times. You must add them as vectors (phasors).

Q: How did we know the phase angles were exactly $45^\circ$?

For both branches, the reactance ($X_C$ or $X_L$) was exactly equal to the resistance ($R$). The phase angle is given by $\tan^{-1}(X/R)$. Since $\tan^{-1}(1) = 45^\circ$, both angles are $45^\circ$.

Q: Why is the voltage across the resistors not just $20\text{ V}$?

In an AC series branch, the total voltage ($20\text{ V}$) is the phasor sum of the voltage across the resistor and the voltage across the reactive component (capacitor or inductor). They don't add up algebraically to $20\text{ V}$, but rather vectorially: $\sqrt{V_R^2 + V_X^2} = 20\text{ V}$.

Q: What does it mean for the inductor and capacitor to be 'ideal'?

An ideal inductor has zero internal resistance, and an ideal capacitor has infinite leakage resistance. This means all the resistance in the branch comes purely from the explicit $100\ \Omega$ and $50\ \Omega$ resistors.

Q: Is the total current in phase with the voltage?

Not necessarily. The total current is the vector sum of $I_1$ and $I_2$. Since $I_2$ (which lags) has a larger magnitude ($0.283\text{ A}$) than $I_1$ ($0.141\text{ A}$), the total current will lag the voltage, making the overall circuit inductive.

Question

In the given circuit, the AC source has $\omega = 100 	ext{ rad/s}$. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

Options:

(A) the current through the circuit, $I$ is approximately $0.3 	ext{ A}$

(B) the current through the circuit, $I$ is approximately $0.3\sqrt{2} 	ext{ A}$

(C) the voltage across $100 \ \Omega$ resistor = $10\sqrt{2} 	ext{ V}$

(D) the voltage across $50 \ \Omega$ resistor = $10 	ext{ V}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When dealing with parallel AC circuits, the most robust approach is to analyze each branch independently before combining the results. In this problem, we are given an AC source of $20	ext{ V}$ operating at an angular frequency $\omega = 100	ext{ rad/s}$. The circuit splits into two parallel branches:
1. **Branch 1 (RC):** A $100\ \mu	ext{F}$ capacitor in series with a $100\ \Omega$ resistor.
2. **Branch 2 (RL):** A $0.5	ext{ H}$ inductor in series with a $50\ \Omega$ resistor.

Our goal is to find the total current $I$ drawn from the source and the voltage drops across specific resistors.

### Evaluating the RC Branch

Let's start with the top branch. First, we calculate the capacitive reactance $X_C$:
$$X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 100 	imes 10^{-6}} = 100\ \Omega$$

Since the resistance $R_1$ is also $100\ \Omega$, the total impedance $Z_1$ of this branch is:
$$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega$$

Because this is an RC circuit, the current $I_1$ will lead the voltage by a phase angle $\phi_1$:
$$\phi_1 = 	an^{-1}\left(\frac{X_C}{R_1}ight) = 	an^{-1}(1) = 45^\circ$$

The magnitude of the current $I_1$ is simply the source voltage divided by the impedance:
$$I_1 = \frac{V}{Z_1} = \frac{20}{100\sqrt{2}} = \frac{1}{5\sqrt{2}}	ext{ A}$$

Now, we can check the voltage across the $100\ \Omega$ resistor. Using Ohm's law:
$$V_{R1} = I_1 R_1 = \left(\frac{1}{5\sqrt{2}}ight) 	imes 100 = 10\sqrt{2}	ext{ V}$$
This perfectly matches option (c)!

### Evaluating the RL Branch

Next, we move to the middle branch. The inductive reactance $X_L$ is:
$$X_L = \omega L = 100 	imes 0.5 = 50\ \Omega$$

With a resistance $R_2 = 50\ \Omega$, the impedance $Z_2$ is:
$$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega$$

For this RL circuit, the current $I_2$ will lag the voltage by a phase angle $\phi_2$:
$$\phi_2 = 	an^{-1}\left(\frac{X_L}{R_2}ight) = 	an^{-1}(1) = 45^\circ$$

The magnitude of the current $I_2$ is:
$$I_2 = \frac{V}{Z_2} = \frac{20}{50\sqrt{2}} = \frac{\sqrt{2}}{5}	ext{ A}$$

Let's check the voltage across the $50\ \Omega$ resistor:
$$V_{R2} = I_2 R_2 = \left(\frac{\sqrt{2}}{5}ight) 	imes 50 = 10\sqrt{2}	ext{ V}$$
Option (d) claims this voltage is $10	ext{ V}$, which is incorrect.

### The Master Calculation: Total Current

To find the total current $I$, we must add $I_1$ and $I_2$ vectorially. We know that $I_1$ leads the voltage by $45^\circ$ and $I_2$ lags the voltage by $45^\circ$. This means the phase difference between the two currents is exactly $90^\circ$!

Because they are perpendicular in the phasor diagram, we can use the Pythagorean theorem to find the magnitude of the total current:
$$I = \sqrt{I_1^2 + I_2^2}$$

$$I = \sqrt{\left(\frac{1}{5\sqrt{2}}ight)^2 + \left(\frac{\sqrt{2}}{5}ight)^2}$$

$$I = \sqrt{\frac{1}{50} + \frac{2}{25}} = \sqrt{\frac{1 + 4}{50}} = \sqrt{\frac{5}{50}} = \frac{1}{\sqrt{10}}	ext{ A}$$

Calculating the decimal value:
$$I \approx \frac{1}{3.162} \approx 0.316	ext{ A}$$
This is approximately $0.3	ext{ A}$, making option (a) correct.

**Final Conclusion:** The correct choices are **(a)** and **(c)**.

Animated Solution for Physics - Electromagnetic Induction: In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

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Visualized Solution

$Circuit Analysis Setup$

$Branch 1: Capacitive Reactance$

$Branch 1: Current and Voltage$

$Branch 2: Inductive Reactance$

$Branch 2: Current and Voltage$

$Total Current via Phasors$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

Evaluating the RC Branch

Evaluating the RL Branch

The Master Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Branch 1: Capacitive Reactance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Branch 1: Current and Voltage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Branch 2: Inductive Reactance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Branch 2: Current and Voltage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Current via Phasors