A series L-C-R circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 kΩ, an in

Q: Why does maximum power occur when $\cos\phi = 1$?

The average power is given by $P_{av} = V_{rms} I_{rms} \cos\phi$. Since $V_{rms}$ and $I_{rms}$ are positive, the power is maximized when the power factor $\cos\phi$ reaches its maximum possible value, which is 1.

Q: What happens to the inductor and capacitor at resonance?

At resonance, the inductive reactance $X_L$ and capacitive reactance $X_C$ are equal in magnitude but opposite in phase. They perfectly cancel each other out, so the circuit behaves as if it only has a resistor.

Q: Why did we ignore the $300\text{ V}$ and $3\text{ k}\Omega$ values in the final calculation?

The condition for maximum power (resonance) only depends on the reactances ($X_L = X_C$). The voltage and resistance determine the *amount* of maximum power, but not the *condition* (the value of $C$) to achieve it.

Q: Can we use $\omega = 2\pi f$ directly?

Yes! The angular frequency $\omega$ is $2\pi f$. So $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. It's just a matter of substituting $f = 50\text{ Hz}$.

Q: What if $\pi^2$ was not given as $10$?

If not given, you would use $\pi \approx 3.14$, so $\pi^2 \approx 9.87$. The approximation $\pi^2 \approx 10$ is very common in physics to simplify manual calculations.

Question

A series L-C-R circuit driven by $300	ext{ V}$ at a frequency of $50	ext{ Hz}$ contains a resistance $R = 3	ext{ k}\Omega$, an inductor of inductive reactance $X_L = 250\pi\Omega$ and an unknown capacitor. The value of capacitance to maximise the average power should be
(Take, $\pi^2 = 10$)

Options:

(A) $4\mu	ext{F}$

(B) $25\mu	ext{F}$

(C) $400\mu	ext{F}$

(D) $40\mu	ext{F}$

Sigma Percentile · Accepted Answer

### Decoding the L-C-R Circuit

Imagine you are an electrical engineer tasked with tuning a circuit to extract the absolute maximum power from an AC source. You are given a series L-C-R circuit connected to a $300	ext{ V}$, $50	ext{ Hz}$ supply.

You know the resistance $R = 3	ext{ k}\Omega$ and the inductive reactance $X_L = 250\pi\,\Omega$. However, the capacitor is missing its label! Your mission is to find the exact capacitance $C$ that makes this circuit hum with maximum power.

### The Secret of Maximum Power

Let's start with the fundamental equation for average power in an AC circuit:

$$P_{av} = V_{rms} I_{rms} \cos\phi$$

Here, $\cos\phi$ is the **power factor**. To maximize the power $P_{av}$, we need this power factor to be as large as possible. Since the maximum value of the cosine function is $1$, we must set:

$$\cos\phi = 1$$

But what does this mean physically? The power factor is defined as the ratio of the true resistance $R$ to the total impedance $Z$ of the circuit:

$$\cos\phi = \frac{R}{Z}$$

If $\frac{R}{Z} = 1$, it immediately tells us that the total impedance must perfectly equal the resistance:

$$Z = R$$

This is the magical state known as **electrical resonance**!

### Unlocking the Resonance Condition

Now, let's look at the full anatomy of the impedance $Z$ in a series L-C-R circuit:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

We substitute this into our resonance condition:

$$\sqrt{R^2 + (X_L - X_C)^2} = R$$

Squaring both sides to eliminate the square root, we get:

$$R^2 + (X_L - X_C)^2 = R^2$$

The $R^2$ terms cancel out beautifully, leaving us with:

$$(X_L - X_C)^2 = 0$$

Which simplifies to our golden rule for resonance:

$$X_L = X_C$$

At resonance, the inductor and capacitor perfectly oppose and cancel each other's effects, leaving the circuit purely resistive!

### Calculating the Unknown Capacitance

We are given the inductive reactance $X_L = 250\pi\,\Omega$. We also know the formula for capacitive reactance:

$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$

Equating the two reactances, we get:

$$250\pi = \frac{1}{2\pi f C}$$

We know the frequency $f = 50	ext{ Hz}$. Let's plug that in:

$$250\pi = \frac{1}{2\pi (50) C}$$

Now, we just need to rearrange the equation to solve for our unknown capacitance $C$:

$$C = \frac{1}{2\pi (50) (250\pi)}$$

Multiplying the terms in the denominator:

$$C = \frac{1}{25000 \pi^2}$$

The problem kindly provides a neat approximation: $\pi^2 = 10$. Substituting this makes the math a breeze:

$$C = \frac{1}{25000 	imes 10} = \frac{1}{250000}$$

Converting this fraction into a decimal:

$$C = 4 	imes 10^{-6}	ext{ F}$$

Which is exactly:

$$C = 4\mu	ext{F}$$

**And there we have it!** By choosing a $4\mu	ext{F}$ capacitor, the circuit enters resonance, the reactances cancel out, and the power dissipation hits its absolute peak. This is the exact same physics that allows you to tune a radio to your favorite station!

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Visualized Solution

$L - C - R Circuit Setup$

$Condition for Maximum Power$

$Power Factor and Impedance$

$Impedance of L-C-R Circuit$

$Resonance Condition$

$Substituting Reactance Values$

$Solving for Capacitance$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Alternating Current (AC) and Voltage

Decoding the L-C-R Circuit

The Secret of Maximum Power

Unlocking the Resonance Condition

Calculating the Unknown Capacitance

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Maximum Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power Factor and Impedance

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Capacitance

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward