### 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.

Letβs say that this is the circuit
weβre 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 πΆ and inductance πΏ of a
circuit. Two ππ equals the square root of
one over πΏπΆ. 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 πΏπΆ.

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 πΆ new equals two
πΆ. The inductance decreasing by
one-half means we are dividing the current inductance by two. So we can write πΏ new equals πΏ
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 πΏ over two two πΆ. 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 πΏπΆ. 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.