# Solve RLC Circuit Problem: Half Power at Resonance

• Saitama
In summary, a series LCR circuit with inductance L = 0.125/π H, capacitance C = 500/π nF and resistance R = 23 Ω is connected to a 230 V variable frequency supply. The power transferred to the circuit at resonance is P = 2300 W. To find the reactance of the circuit when the power transferred is half of P, the formula P' = V^2_rms/Z*cos(φ) is used. Solving for the reactance X, it is found to be 23 Ω, not 23√2 Ω as initially calculated.
Saitama

## Homework Statement

A series LCR circuit with ##L=0.125/\pi## H, ##C=500/\pi## nF and ##R=23\,\Omega## is connected to a 230 V variable frequency supply. For what reactance of circuit, the power transferred to the circuit is half the power at resonance?

## The Attempt at a Solution

At resonance,
$$f=\frac{1}{2\pi\sqrt{LC}}=2000\,Hz$$
Hence, the power transferred at resonance is given by ##P=V^2_{rms}/R=2300\,\,W##.

When the power transferred is half, let the reactance be Z, hence,
$$P'=\frac{V^2_{rms}}{Z}\cos\phi=\frac{V^2_{rms}}{Z}\frac{R}{Z}$$
As per the question:
$$\frac{V^2_{rms}}{Z}\frac{R}{Z}=\frac{1}{2}\times 2300$$
$$\Rightarrow \frac{230\times 230 \times 23}{Z^2}=\frac{1}{2}\times 2300$$
$$\Rightarrow Z=23\sqrt{2} \,\,\Omega$$
But this is incorrect. The correct answer is ##23\,\,\Omega##.

Any help is appreciated. Thanks!

The way you've defined it, Z is the impedance, not the reactance.

The reactance, X is such that Z = R + jX, and it's the X that you need to solve for.

Note that the magnitude squared of Z is, Z2 = R2 + X2. That will come in useful.

Last edited:
1 person
collinsmark said:
The way you've defined it, Z is the impedance, not the reactance.

Ah yes, thanks a lot collinsmark!

## 1. What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor, inductor, and capacitor. These components are connected in series or parallel and their interactions create a resonant frequency.

## 2. How do you calculate the resonant frequency for an RLC circuit?

The resonant frequency can be calculated using the formula: f = 1 / (2π√(LC)), where f is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.

## 3. What does it mean for a circuit to have half power at resonance?

Half power at resonance means that the power dissipated in the circuit is half of its maximum value. This occurs when the impedance of the circuit is equal to the resistance, resulting in a voltage drop across the resistor that is half of the input voltage.

## 4. How do you solve for half power at resonance?

To solve for half power at resonance, you can use the formula: P = (Vin)2 / (4R), where P is the power, Vin is the input voltage, and R is the resistance. Set this equal to half of the maximum power and solve for the resonant frequency using the formula mentioned in question 2.

## 5. What are some real-life applications of RLC circuits?

RLC circuits have many practical applications, including in radio and television receivers, electronic filters, electric power transmission and distribution systems, and audio speakers. They are also used in electronic devices such as computers, cell phones, and electric motors.

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