In an L-C-R series circuit, an inductor 30 mH and a resistor 1 ohm are connected to an AC source of

Q: Why is the formula $X_C - X_L$ and not $X_L - X_C$?

Because the question states that the current leads the voltage. This means the circuit is predominantly capacitive, so the capacitive reactance $X_C$ must be greater than the inductive reactance $X_L$.

Q: What does it physically mean when current leads voltage?

It means the capacitor's effect dominates the inductor's effect. The capacitor charges and discharges in a way that pulls the current ahead of the driving voltage.

Q: How did we get $\tan 45^\circ = 1$?

This is a standard trigonometric value. A $45^\circ$ angle in a right triangle means the opposite and adjacent sides are equal in length.

Q: Why did we convert 30 mH to $30 \times 10^{-3}$ H?

To ensure all our calculations are in standard SI units. Mixing millihenries with ohms and radians per second will lead to an incorrect final answer.

Q: What if the phase angle was $0^\circ$?

If the phase angle was $0^\circ$, it would mean $X_C = X_L$. This is the condition for resonance, where the circuit behaves purely resistively.

Question

In an L-C-R series circuit, an inductor $30 	ext{ mH}$ and a resistor $1 \Omega$ are connected to an AC source of angular frequency $300 	ext{ rad/s}$. The value of capacitance for which, the current leads the voltage by $45^\circ$ is $\frac{1}{x} 	imes 10^{-3} 	ext{ F}$. Then, the value of x is …… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a classic L-C-R series circuit. We have an inductor with $L = 30 	ext{ mH}$, a resistor with $R = 1 \Omega$, and an unknown capacitor $C$, all driven by an AC source with an angular frequency $\omega = 300 	ext{ rad/s}$.

The most crucial piece of information given is that the **current leads the voltage by $45^\circ$**. In the world of AC circuits, this is a massive clue! It tells us that the circuit is predominantly capacitive, meaning the capacitive reactance ($X_C$) is overpowering the inductive reactance ($X_L$).

### The Master Equation

To mathematically capture this relationship, we turn to the impedance triangle. The phase angle $\phi$ is related to the reactances and resistance by the formula:
$$ 	an \phi = \frac{X_C - X_L}{R} $$

Since we know $\phi = 45^\circ$ and $R = 1 \Omega$, we can substitute these right in. And because $	an 45^\circ = 1$, our equation simplifies beautifully:
$$ 1 = \frac{X_C - X_L}{1} \implies X_C - X_L = 1 $$

### Unlocking the Reactances

Now, let's find out what $X_L$ is. The inductive reactance is simply the product of the angular frequency and the inductance:
$$ X_L = \omega L = 300 	imes 30 	imes 10^{-3} = 9 \Omega $$

Armed with $X_L$, we can easily find $X_C$ from our simplified master equation:
$$ X_C - 9 = 1 \implies X_C = 10 \Omega $$

### Final Calculation

We are in the endgame now. The capacitive reactance $X_C$ is inversely proportional to the capacitance $C$:
$$ X_C = \frac{1}{\omega C} $$

Substituting our known values:
$$ 10 = \frac{1}{300 	imes C} $$

Rearranging for $C$, we get:
$$ C = \frac{1}{3000} 	ext{ F} $$

To match the format requested in the question, we can rewrite this as:
$$ C = \frac{1}{3} 	imes 10^{-3} 	ext{ F} $$

Comparing this to the given expression $\frac{1}{x} 	imes 10^{-3} 	ext{ F}$, it is crystal clear that:
$$ x = 3 $$

Enter Numerical Value:

Visualized Solution

$L, C, R Circuit$

$tan ϕ = \frac{X _{C} - X _{L}}{R}$

$ϕ = 4 5^{\circ}$

$X_{L} = ω L$

$X_{C} = X_{L} + R$

$C = \frac{1}{ω X _{C}}$

$x = 3$

$X_{L} > X_{C} \Rightarrow Lag$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Unlocking the Reactances

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanϕ=RXC​−XL​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ=45∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XL​=ωL

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XC​=XL​+R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=ωXC​1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XL​>XC​⇒Lag