Question

in this problem we consider the resonance curve for a circuit with electrical components l, r, and c and resonant frequency ω₀
part a
for given values of r and c, if you double the value of l, how does the new resonance curve differ from the original one?
view available hint(s)
both the peak height and peak frequency double.
the peak height will be half as great, and the peak frequency will double.
the peak height wont change, and the peak frequency will be 1/√2 times as great.
the peak height wont change, and the peak frequency will be half as great.
submit
part b

Explanation:

Step1: Recall the formula for resonant frequency

The resonant frequency formula is $\omega_0=\frac{1}{\sqrt{LC}}$.

Step2: Analyze the effect of doubling \(L\) on \(\omega_0\)

Let the original \(L\) be \(L_1\) and new \(L\) be \(L_2 = 2L_1\). The original \(\omega_{01}=\frac{1}{\sqrt{L_1C}}\), and the new \(\omega_{02}=\frac{1}{\sqrt{L_2C}}=\frac{1}{\sqrt{2L_1C}}=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{L_1C}}=\frac{\omega_{01}}{\sqrt{2}}\).

Step3: Recall the formula for peak height

The peak height of the resonance curve (at resonance) is \(I_0=\frac{V}{R}\) (from \(I = \frac{V}{Z}\) and at resonance \(Z = R\)). Since \(V\) and \(R\) are unchanged, the peak height \(I_0\) does not change.

Answer:

The peak height won't change, and the peak frequency will be \(1/\sqrt{2}\) times as great.