In a series L-R circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor

Q: Why do we use $V_{rms}$ instead of peak voltage?

AC power formulas inherently use RMS values because they represent the equivalent DC voltage that would deliver the same average power.

Q: What does a power factor of unity physically mean?

It means the voltage and current are perfectly in phase. The circuit behaves as if it only has resistance, maximizing the active power delivered to the load.

Q: Why did we add the capacitor in series and not parallel?

Adding it in series directly subtracts its reactance from the inductive reactance ($X = X_L - X_C$). If added in parallel, the analysis would involve admittances, which is a different approach for power factor correction.

Q: How did we know the initial circuit was just L and R?

The problem states 'In a series L-R circuit', explicitly telling us there is no capacitor initially.

Q: Why did we convert Farads to microfarads at the end?

The question specifically asked for the value of $n$ in the expression $\left(\frac{n}{3\pi}\right)\mu\text{F}$. Matching the units is crucial to extracting the correct integer value.

Question

In a series $L-R$ circuit, power of $400	ext{W}$ is dissipated from a source of $250	ext{ V}$, $50	ext{ Hz}$. The power factor of the circuit is $0.8$. In order to bring the power factor to unity, a capacitor of value $C$ is added in series to the $L$ and $R$. Taking the value of $C$ as $\left(\frac{n}{3\pi}ight)\mu	ext{F}$, then value of $n$ is ……… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are analyzing an alternating current circuit. We have an inductor and a resistor connected in series to a $250	ext{ V}$, $50	ext{ Hz}$ AC source.

The power dissipated in this initial setup is $400	ext{ W}$, and the power factor is $0.8$.

Our mission is to find the exact capacitance needed to bring this power factor to unity. This is a classic problem of power factor correction, a crucial concept in electrical engineering.

### The Master Equation

Let's start with the power equation. The average power dissipated in an AC circuit is given by the product of the RMS voltage, the RMS current, and the power factor.

$$P = V_{rms} I_{rms} \cos\phi$$

Since the RMS current is simply the RMS voltage divided by the total impedance $Z$, we can rewrite the power formula.

$$P = \frac{V_{rms}^2}{Z} \cos\phi$$

Now, let's rearrange this equation to solve for the impedance $Z$. We bring $Z$ to the left and power $P$ to the denominator on the right.

$$Z = \frac{V_{rms}^2}{P} \cos\phi$$

Substituting the values we know, we can set up the calculation.

$$Z = \frac{(250)^2}{400} 	imes 0.8$$

Let's compute this step by step. Squaring $250$ gives $62500$. Dividing that by $400$ gives $156.25$. Finally, multiplying by $0.8$ yields an impedance $Z$ of exactly **$125\ \Omega$**. This is the total opposition to current in our initial L-R circuit.

### Finding the Resistance and Reactance

Next, we need to find the pure resistance $R$. We know from the impedance triangle that the power factor, $\cos\phi$, is equal to the resistance $R$ divided by the impedance $Z$.

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

Plugging in $0.8$ for the power factor and $125$ for $Z$, we can easily solve for $R$.

$$0.8 = \frac{R}{125} \implies R = 100\ \Omega$$

With $Z$ and $R$ known, we can determine the inductive reactance $X_L$. We use the Pythagorean relation for impedance.

$$Z^2 = R^2 + X_L^2$$

So, $X_L$ is the square root of $125^2$ minus $100^2$. This is a classic 3-4-5 right triangle scaled up by a factor of 25.

$$X_L = \sqrt{125^2 - 100^2} = 75\ \Omega$$

### The Resonance Condition

Here is the crucial conceptual leap. The problem asks us to bring the power factor to unity, which means $\cos\phi = 1$.

For this to happen, the circuit must behave as if it is purely resistive. This occurs at **resonance**, where the capacitive reactance $X_C$ perfectly cancels the inductive reactance $X_L$.

$$X_C = X_L = 75\ \Omega$$

### Final Calculation

Now, let's find the actual capacitance. We know $X_C$ is inversely proportional to the angular frequency and capacitance.

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

Substituting $50	ext{ Hz}$ for $f$ and $75\ \Omega$ for $X_C$, we can solve for $C$.

$$75 = \frac{1}{2\pi (50) C} \implies C = \frac{1}{7500\pi}	ext{ F}$$

To match the format given in the question, we need to convert Farads to microfarads by multiplying by $10^6$.

$$C = \frac{10^6}{7500\pi}\ \mu	ext{F} = \frac{10000}{75\pi}\ \mu	ext{F} = \frac{400}{3\pi}\ \mu	ext{F}$$

Comparing this with the given expression $\left(\frac{n}{3\pi}ight)\mu	ext{F}$, we can confidently say that $n$ equals **$400$**. What a brilliant application of AC circuit principles!

Enter Numerical Value:

Visualized Solution

$L - R - C Circuit$

$Power Equation$

$Impedance Setup$

$Calculating Z$

$Finding Resistance R$

$Inductive Reactance X_{L}$

$Unity Power Factor$

$Calculating Capacitance C$

$Final Answer$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Finding the Resistance and Reactance

The Resonance Condition

Final Calculation

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Comprehension Passage

The value of $C$ is _______.

The value of $ϕ$ is _______.

An AC circuit has $R = 100Ω$ , $C = 2 μ F$ and $L = 80 mH$ connected in series. The quality factor of the circuit is

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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−R−C Circuit

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

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

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

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

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

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

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

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