Question

An ideal inductor, a resistor of resistance \(R\) Ohms and a capacitor with adjustable capacitance are connected in series to an alternating voltage with an effective value of \(V\) Volts and frequency of \(f\) Hz. The current flowing through the circuit when the capacitance of the capacitor is set to \(C_1\) is the same as when the capacitance of the capacitor is set to \(C_2\), \(C_2 > C_1\). The inductance of the inductor \(L\) is given by:

• A. \(\dfrac{1}{8\pi^2 f^2}\dfrac{C_1 + C_2}{C_1 C_2}\)
• B. \(\dfrac{1}{8\pi^2 f^2}\dfrac{C_1 C_2}{C_1 + C_2}\)
• C. \(\dfrac{1}{8\pi^2 f^2}\dfrac{C_1 - C_2}{C_1 C_2}\)
• D. \(\dfrac{1}{2\pi^2 f^2}\dfrac{1}{R(C_1 - C_2)}\dfrac{C_1 + C_2}{C_1 C_2}\)

Hint:

Equal current means the two settings are symmetric about resonance: \(X_L - X_{C_1} = -(X_L - X_{C_2})\), so \(2\cdot 2\pi f L = \tfrac{1}{2\pi f C_1} + \tfrac{1}{2\pi f C_2}\).

Answer 1

The Correct Option is A

Step 1: In a series RLC circuit the current amplitude is \[I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}},\] where \(X_L = 2\pi f L\) and \(X_C = \dfrac{1}{2\pi f C}\).
Step 2: The same current at two different capacitances means the impedance magnitude is the same. Since \(R\) and \(X_L\) do not change, we need \[(X_L - X_{C_1})^2 = (X_L - X_{C_2})^2.\]
Step 3: Because \(C_2 > C_1\), we have \(X_{C_2} < X_{C_1}\). The only non-trivial solution is that the circuit is inductive in one case and capacitive in the other, with equal magnitudes: \[X_L - X_{C_2} = -(X_L - X_{C_1}).\]
Step 4: Rearranging gives \[2X_L = X_{C_1} + X_{C_2}.\] Substitute the reactances: \[2(2\pi f L) = \frac{1}{2\pi f C_1} + \frac{1}{2\pi f C_2}.\]
Step 5: Multiply out. The right side is \[\frac{1}{2\pi f}\left(\frac{1}{C_1} + \frac{1}{C_2}\right) = \frac{1}{2\pi f}\cdot\frac{C_1 + C_2}{C_1 C_2}.\] So \[4\pi f L = \frac{1}{2\pi f}\cdot\frac{C_1 + C_2}{C_1 C_2}.\]
Step 6: Solve for \(L\): \[L = \frac{1}{8\pi^2 f^2}\cdot\frac{C_1 + C_2}{C_1 C_2}.\] \[\boxed{L = \frac{1}{8\pi^2 f^2}\,\frac{C_1 + C_2}{C_1 C_2}}\]