Question Video: Using the 𝑄 Factor to Calculate Capacitance Physics • Third Year of Secondary School

Video Transcript

A circuit containing a resistor, a capacitor, and an inductor is used as a receiver for electromagnetic waves with a resonant frequency of 121 kilohertz. The resistance of the resistor is 116 kiloohms. The circuit has a 𝑄 factor of 1.50. What is the capacitance of the capacitor in the circuit? The formula used to calculate the 𝑄 factor is 𝑄 equals one over 𝑅 multiplied by the square root of 𝐿 over 𝐶. Give your answer in scientific notation to two decimal places.

In this question, we are given an RLC circuit, and we have to determine the capacitance using the 𝑄 factor, which we are told is equal to 1.50. Let’s say that this is the circuit we’re working with. It has a resistor, an inductor, and a capacitor. And since this is an alternating current circuit, it has a variable voltage supply. We are given the resonant frequency 𝑓 of the circuit and the resistance of the resistor 𝑅. But we do not know the inductance 𝐿 of the circuit.

To determine the inductance, we will recall an equation that can relate 𝐿 to our other variables. The resonant frequency 𝑓 of a circuit is given by the equation two 𝜋𝑓 equals the square root of one over 𝐿𝐶, where 𝐿 is the inductance of the circuit and 𝐶 is the capacitance of the circuit.

We now need to make 𝐿 the subject. We can do this by squaring both sides, which gives two 𝜋𝑓 squared equals one over 𝐿𝐶. We can then take the reciprocal of both sides to give us one over two 𝜋𝑓 squared equals 𝐿𝐶. And finally, we can divide both sides by 𝐶 to leave us with 𝐿 equals one over two 𝜋𝑓 squared 𝐶. Then, we substitute this equation for 𝐿 into the equation for the 𝑄 factor to get 𝑄 equals one over 𝑅 multiplied by the square root of one over two 𝜋𝑓 squared 𝐶 squared. The square root will cancel the squares in the bottom terms to give 𝑄 equals one over two 𝜋𝑓𝐶𝑅.

We want to calculate the capacitance of the capacitor in the circuit. So we need to rearrange this equation to make 𝐶 the subject. We can do this by multiplying both sides of the equation by 𝐶 and dividing both sides by 𝑄 to give us 𝐶 equals one over two 𝜋𝑓𝑅𝑄.

Before we substitute in the given variables, we need to be careful of the unit prefixes. The resonant frequency is given as 121 kilohertz, which is equal to 121 times 10 to the power three hertz. The resistance 𝑅 is given as 116 kiloohms, which is equal to 116 times 10 to the power three ohms. And the 𝑄 factor is given as 1.50 and is a dimensionless number.

Now then, if we substitute in our given variables into this equation, we find that 𝐶 is equal to one over two 𝜋 multiplied by 121 times 10 to the power three hertz multiplied by 116 times 10 to the power three ohms multiplied by 1.50. Completing this calculation, we find that the capacitance 𝐶 in scientific notation to two decimal places is equal to 7.56 times 10 to the power negative 12 farads. And this is our answer.

The capacitance of the capacitor in the circuit is equal to 7.56 times 10 to the power negative 12 farads.