Question Video: Calculating the Q Factor of an RLC Circuit | Nagwa Question Video: Calculating the Q Factor of an RLC Circuit | Nagwa

# Question Video: Calculating the Q Factor of an RLC Circuit Physics • Third Year of Secondary School

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If the π factor of a RLC circuit is calculated using the formula π = (1/π) β(πΏ/πΆ), calculate the π factor of a circuit that contains a 555 mH inductor and a 32.4 kΞ© resistor if the resonant frequency of the circuit is 247 kHz. Give your answer to one decimal place.

03:28

### Video Transcript

If the π factor of a RLC circuit is calculated using the formula π equals one over π square root πΏ over πΆ, calculate the π factor of a circuit that contains a 555-millihenry inductor and a 32.4-kiloohm resistor if the resonant frequency of the circuit is 247 kilohertz. Give your answer to one decimal place.

In this question, we are asked to calculate π factor of a RLC circuit using the equation given to us.

Letβs say that this is the circuit weβre working with. It has a resistor, inductor, and capacitor. And since this is an alternating current circuit, it has a variable voltage supply. We are given the resistance π of the resistor and the inductance πΏ of the inductor. But the value of the capacitance πΆ is not stated in the question. To determine the capacitance, we will recall an equation that can relate πΆ to our other variables.

The resonant frequency π of a circuit is given by the equation two ππ equals the square root of one over πΏπΆ, where πΏ is the inductance of the circuit and πΆ is the capacitance of the circuit. We now need to make πΆ the subject. We can do this by squaring both sides, which gives two ππ squared equals one over πΏπΆ. We can then take the reciprocal of both sides to give us one over two ππ squared equals πΏπΆ. And finally, we can divide both sides by πΏ to leave us with πΆ equals one over two ππ squared πΏ.

We can now substitute this equation for the capacitance πΆ into the equation of the π factor that we were given in the question to get π equals one over π multiplied by the square root of two ππ squared πΏ squared. The square root will cancel the squared terms, leaving us with π equals two πππΏ over π.

Before we substitute in the given variables, we need to be careful of the unit prefixes. The resonant frequency is given as 247 kilohertz, which is equal to 247 times 10 to the power three hertz. The inductance πΏ is given as 555 millihenries, which is equal to 555 times 10 to the power negative three henries. And the resistance π is given as 32.4 kiloohms, which is equal to 32.4 times 10 to the power three ohms.

Now then, if we substitute in our given variables into this equation, we find that π is equal to two π multiplied by 247 times 10 to the power three hertz multiplied by 555 times 10 to the power negative three henries over 32.4 times 10 to the power three ohms. Completing this calculation, we find that π is equal to 26.6 when rounded to one decimal place, which is a dimensionless number.

So, the π factor for RLC circuit with a 555-millihenry inductor, a 32.4-kiloohm resistor, and a resonant frequency of 247 kilohertz is 26.6.

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