In L-C circuit, the inductance L = 40 mH and capacitance C = 100 uF. If a voltage V(t) = 10 sin(314

Q: Why do we subtract $X_L$ from $X_C$ to find the impedance?

In an AC circuit, the voltage across an inductor leads the current by $90^\circ$, while the voltage across a capacitor lags the current by $90^\circ$. This means their voltages (and thus their reactances) are exactly $180^\circ$ out of phase. They directly oppose each other, so we subtract them to find the net effect.

Q: How do I know if the current leads or lags the voltage?

You look at which reactance is larger. If $X_C > X_L$, the circuit is capacitive, and current leads voltage. If $X_L > X_C$, the circuit is inductive, and current lags voltage.

Q: Why did the sine function change to a cosine function in the final answer?

Because the circuit is capacitive, the current leads the voltage by $90^\circ$ (or $\frac{\pi}{2}$ radians). Mathematically, adding $\frac{\pi}{2}$ to a sine function transforms it into a cosine function: $\sin(\theta + \frac{\pi}{2}) = \cos\theta$.

Q: Why is there no resistance $R$ in the impedance formula?

The problem specifically states it is an $L-C$ circuit, and no resistor value is given. We assume it is an ideal circuit where the resistance $R = 0$. Therefore, the impedance formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$ simplifies to just $Z = |X_L - X_C|$.

Q: What would happen if $X_L$ and $X_C$ were exactly equal?

If $X_L = X_C$, their effects would completely cancel out, making the net impedance $Z = 0$ (since $R=0$). This is the condition for electrical resonance. In a purely ideal $L-C$ circuit, the current would theoretically become infinite!

Question

In $L-C$ circuit, the inductance $L = 40$ mH and capacitance $C = 100 \mu	ext{F}$. If a voltage $V(t) = 10 \sin(314 t)$ is applied to the circuit, the current in the circuit is given as

Options:

(A) $0.52 \cos 314 t$

(B) $0.52 \sin 314 t$

(C) $10 \cos 314 t$

(D) $5.2 \cos 314 t$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a simple yet fascinating electrical circuit. We have an inductor with an inductance of $L = 40 	ext{ mH}$ and a capacitor with a capacitance of $C = 100 \mu	ext{F}$ connected in series. Powering this setup is an alternating voltage source described by the equation $V(t) = 10 \sin(314 t)$.

From this voltage equation, we can immediately extract two critical pieces of information. The peak voltage, $V_m$, is simply the amplitude of the sine wave, which is $10 	ext{ V}$. The angular frequency, $\omega$, is the coefficient of $t$ inside the sine function, giving us $\omega = 314 	ext{ rad/s}$.

### Calculating Reactances

In an AC circuit, both inductors and capacitors offer opposition to the flow of alternating current. This opposition is known as reactance. Let's calculate the reactance for each component.

First, the inductive reactance, $X_L$, is directly proportional to the frequency and the inductance. The formula is $X_L = \omega L$. Substituting our values, we get:

$$X_L = 314 	imes 40 	imes 10^{-3} = 12.56 \, \Omega$$

Next, we calculate the capacitive reactance, $X_C$. Unlike the inductor, a capacitor's opposition is inversely proportional to the frequency and capacitance. The formula is $X_C = \frac{1}{\omega C}$. Plugging in the numbers:

$$X_C = \frac{1}{314 	imes 100 	imes 10^{-6}} \approx 31.85 \, \Omega$$

### The Net Impedance and Phase

Because the inductor and capacitor are connected in series, their reactances oppose each other. They are exactly $180^\circ$ out of phase. Therefore, the net impedance, $Z$, of this $L-C$ circuit is simply the absolute difference between the two reactances:

$$Z = |X_C - X_L| = 31.85 - 12.56 = 19.29 \, \Omega$$

Here is the crucial catch: notice that $X_C > X_L$. Because the capacitive reactance is greater, the entire circuit behaves effectively like a purely capacitive circuit. In a purely capacitive circuit, the current leads the voltage by exactly $90^\circ$, or $\frac{\pi}{2}$ radians.

### Final Current Equation

Now that we have the net impedance, finding the peak current, $I_m$, is straightforward using Ohm's law for AC circuits:

$$I_m = \frac{V_m}{Z} = \frac{10}{19.29} \approx 0.52 	ext{ A}$$

Finally, we construct the equation for the instantaneous current, $I(t)$. We know the peak current is $0.52 	ext{ A}$, and we established that the current leads the voltage by $\frac{\pi}{2}$. Therefore, we add $\frac{\pi}{2}$ to the phase of the voltage equation:

$$I(t) = 0.52 \sin\left(314 t + \frac{\pi}{2}ight)$$

Using the trigonometric identity $\sin\left(	heta + \frac{\pi}{2}ight) = \cos	heta$, we arrive at our final, elegant answer:

**$$I(t) = 0.52 \cos(314 t)$$**

Animated Solution for Physics - Electromagnetic Induction: In $L - C$ circuit, the inductance $L = 40$ mH and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 sin (314 t)$ is applied to the circuit, the current in the circuit is given as

Select Answer:

Visualized Solution

$L - C$ Circuit Setup

Voltage Equation Analysis

Inductive Reactance ( $X_{L}$ )

Capacitive Reactance ( $X_{C}$ )

Net Impedance ( $Z$ )

Phase Angle

Peak Current ( $I_{m}$ )

Final Current Equation

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

Calculating Reactances

The Net Impedance and Phase

Final Current Equation

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Select Answer:

Visualized Solution

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

Voltage Equation Analysis

Inductive Reactance ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XL​)

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

Net Impedance ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z)

Phase Angle

Peak Current ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Im​)

Final Current Equation

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

Calculating Reactances

The Net Impedance and Phase

Final Current Equation

Similar Questions

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

Inductive Reactance ( $X_{L}$ )

Capacitive Reactance ( $X_{C}$ )

Net Impedance ( $Z$ )

Peak Current ( $I_{m}$ )

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)

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an AC source $V = V_{0} sin ω t$ as shown. Which of the following options is/are correct?