Question Video: Calculating Changes in Resonant Frequency | Nagwa Question Video: Calculating Changes in Resonant Frequency | Nagwa

# Question Video: Calculating Changes in Resonant Frequency Physics • Third Year of Secondary School

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For an RLC circuit, if the total capacitance of the circuit is doubled and the total inductance of the circuit is decreased to its half, what will happen to the resonant frequency of the circuit?

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### Video Transcript

For an RLC circuit, if the total capacitance of the circuit is doubled and the total inductance of the circuit is decreased to its half, what will happen to the resonant frequency of the circuit? (A) It will increase to four times its value. (B) It will decrease to its quarter. (C) It will stay the same. (D) It will decrease to its half.

In this question, there is an RLC circuit. And we are asked to determine what would happen to the resonant frequency of the circuit if the total capacitance of the circuit is doubled and the total inductance of the circuit is decreased by half.

Letβs say that this is the circuit weβ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 can recall a formula that relates the resonant frequency π to the capacitance πΆ and inductance πΏ of a circuit. Two ππ equals the square root of one over πΏπΆ. Since we are interested in how the resonant frequency will change, we can make π the subject by simply dividing both sides of the equation by two π, leaving us with π equals one over two π multiplied by the square root of one over πΏπΆ.

We now have an equation that links the resonant frequency to the capacitance and inductance of this circuit. Now we want to figure out what would happen if the total capacitance in this circuit is doubled and the total inductance of the circuit is decreased to its half. The capacitance doubling means we are multiplying the current capacitance by two, so we can write πΆ new equals two πΆ. The inductance decreasing by one-half means we are dividing the current inductance by two. So we can write πΏ new equals πΏ over two.

We can now substitute these values into the equation for the resonant frequency of the circuit to see how the resonant frequency will change. We find that the new resonant frequency π new is equal to one over two π multiplied by the square root of one over πΏ over two two πΆ. When we do this, we can cancel out the two that is multiplied by the capacitance with the two that is under the inductance, leaving us with one over two π multiplied by the square root of one over πΏπΆ. This formula is equal to the old resonant frequency, π. So we have found that there will be no change to the resonant frequency with these changes to the capacitance and inductance of the circuit.

Looking through our options, we can see that option (C) states that the resonant frequency will remain the same, which is what we found when applying the situation stated in the problem to the equation for resonant frequency. Therefore, option (C), it will stay the same, is the correct answer.

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