A sinusoidal voltage of peak value 250 V is applied to a series L-C-R circuit, in which R = 8 ohm, L

Q: Why didn't we use the values of L and C in the calculation?

At resonance, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) are exactly equal and opposite. They cancel each other out completely, meaning the circuit's total impedance is just the resistance ($R$). The specific values of L and C determine *when* resonance happens, but at that exact moment, they don't consume any net real power!

Q: Why do we use $V_{rms}$ instead of $V_0$ for power?

In AC circuits, power fluctuates constantly. To find the *average* power dissipated over a full cycle, we use RMS (Root Mean Square) values. Using the peak voltage $V_0$ directly would give us the maximum instantaneous power, not the average power.

Q: What if the circuit wasn't at resonance?

If it wasn't at resonance, we would have to calculate the total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and the power factor $\cos\phi = R/Z$. The power would then be $P = V_{rms} I_{rms} \cos\phi$. It would be a much longer calculation!

Q: Why is the power factor 1 at resonance?

The power factor is $\cos\phi$, where $\phi$ is the phase angle between voltage and current. At resonance, the circuit is purely resistive, meaning voltage and current are perfectly in phase ($\phi = 0^\circ$). Since $\cos(0^\circ) = 1$, the power factor is 1.

Q: I got 3906 W, why is the answer 4?

The question specifically asks for the power in **kilowatts (kW)** and to the **nearest integer**. 3906.25 W is 3.90625 kW. Rounding 3.90625 to the nearest whole number gives us 4.

Question

A sinusoidal voltage of peak value $250	ext{ V}$ is applied to a series L-C-R circuit, in which $R = 8\,\Omega$, $L = 24	ext{ mH}$ and $C = 60\,\mu	ext{F}$. The value of power dissipated at resonant condition is $x	ext{ kW}$. The value of $x$ to the nearest integer is ...... .

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

Imagine you are staring at a dense jungle of electrical components: an inductor, a capacitor, a resistor, and an AC voltage source all wired together in series. The problem throws a bunch of numbers at you—$24 	ext{ mH}$ for inductance, $60 \mu	ext{F}$ for capacitance. It looks intimidating, right?

But here is the secret: physics often hides elegant shortcuts in plain sight. The magic word in this problem is **"resonant condition"**.

When an L-C-R circuit hits resonance, something beautiful happens. The inductive reactance ($X_L$) and the capacitive reactance ($X_C$) become perfectly equal. Because they act in opposite directions, they completely cancel each other out.

### The Magic of Resonance

What does this cancellation mean for our circuit? It means the circuit forgets about the inductor and the capacitor entirely! The total impedance ($Z$) of the circuit drops to its absolute minimum, which is simply the resistance ($R$).

So, $Z = R = 8 \Omega$.

This is a massive relief. We don't need to calculate complex impedances or worry about phase angles. The circuit behaves exactly like a pure resistive circuit. In a pure resistive circuit, the voltage and current are perfectly in phase, meaning the **power factor ($\cos\phi$) is exactly 1**.

### The Power Equation

Now, let's talk about power. The general formula for average power dissipated in an AC circuit is $P = V_{rms} I_{rms} \cos\phi$.

Since we are at resonance, $\cos\phi = 1$. And from Ohm's law, the RMS current is simply the RMS voltage divided by the resistance ($I_{rms} = V_{rms} / R$).

Substituting this into our power equation gives us a beautifully simple formula: 
$$P = \frac{V_{rms}^2}{R}$$

But wait, there is a trap! The question gives us the **peak voltage** ($V_0 = 250 	ext{ V}$), not the RMS voltage. We must convert it before plugging it into our formula.

The relationship is $V_{rms} = \frac{V_0}{\sqrt{2}}$.

So, our RMS voltage is $\frac{250}{\sqrt{2}} 	ext{ V}$.

### The Final Calculation

Let's bring it all together and substitute our values into the power formula:

$$P = \frac{\left(\frac{250}{\sqrt{2}}ight)^2}{8}$$

First, we square the numerator. The square of $250$ is $62500$, and the square of $\sqrt{2}$ is simply $2$.

$$P = \frac{\frac{62500}{2}}{8}$$

Dividing $62500$ by $2$ gives us $31250$. Now we divide by the resistance, $8 \Omega$:

$$P = \frac{31250}{8} = 3906.25 	ext{ W}$$

We have our power in watts, but the question asks for the answer in **kilowatts (kW)**. To convert, we divide by $1000$:

$$P = 3.90625 	ext{ kW}$$

Finally, the question asks for the value of $x$ to the **nearest integer**. Looking at $3.90625$, the nearest whole number is $4$.

Therefore, our final answer is $x = 4$.

The extra values of inductance and capacitance were just there to test your conceptual clarity. Once you spot the word "resonance", the problem becomes a breeze!

Enter Numerical Value:

Visualized Solution

$L - C - R Circuit Setup$

$Resonance Condition$

$Power Dissipated$

$RMS Voltage$

$Substitution$

$Calculation$

$Final Power$

$Nearest Integer$

$The Way Forward$

The Sigma Insight: Alternating Current (AC) and Voltage

The Illusion of Complexity

The Magic of Resonance

The Power Equation

The Final Calculation

Similar Questions

A series L-C-R circuit of $R = 5Ω$ , $L = 20 mH$ and $C = 0.5 μ F$ is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ...... $\times 1 0^{2} W$ .

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

In a series resonant $L$ - $C$ - $R$ circuit, the voltage across $R$ is $100 V$ and $R = 1 k Ω$ with $C = 2 μ F$ . The resonant frequency $ω$ is $200 rad/s$ . At resonance, the voltage across $L$ is

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In an L-C-R series circuit, an inductor $30 mH$ and a resistor $1Ω$ are connected to an AC source of angular frequency $300 rad/s$ . The value of capacitance for which, the current leads the voltage by $4 5^{\circ}$ is $\frac{1}{x} \times 1 0^{- 3} F$ . Then, the value of x is …… .

In a series $L - C - R$ circuit, the inductive reactance ( $X_{L}$ ) is $10 Ω$ and the capacitive reactance ( $X_{C}$ ) is $4 Ω$ . The resistance ( $R$ ) in the circuit is $6 Ω$ . The power factor of the circuit is

An AC circuit has an inductor and a resistor of resistance $R$ in series, such that $X_{L} = 3 R$ . Now, a capacitor is added in series such that $X_{C} = 2 R$ . The ratio of new power factor with the old power factor of the circuit is $5 : x$ . The value of $x$ is.

A $100 Ω$ resistance, a $0.1 μ F$ capacitor and an inductor are connected in series across a $250 V$ supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is $60 Hz$ .

Two circuits are shown in the figures (a) and (b). At a frequency of ......... rad/s, the average power dissipated in one cycle will be same in both the circuits.

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 Setup

.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; } Power Dissipated

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

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

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

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

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

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