Question

When an ac source with a peak potential difference of 40 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is C=50μF, the frequency is 130 Hz, and the current leads the potential difference by 300 . Find the resitance and the inductance.

Answer 1

The resistance of the series RLC circuit is 20 ohms and the inductance is 0.000118 H (or 118 µH).

Step-by-step explanation:

To find the resistance and inductance of the series RLC circuit, we can use the given information and equations related to the circuit's properties.

The peak potential difference (Vp) is 40 V and the peak current (Ip) is 2 A. The capacitor (C) is 50µF, the frequency (f) is 130 Hz, and the current leads the potential difference by 300°.

First, we can calculate the resistance (R) using Ohm's Law, R = Vp / Ip. Substituting the given values, R = 40 V / 2 A = 20 ohms.

To find the inductance (L), we can use the formula XL = Vp / (Ip * 2 * π * f), where XL is the inductive reactance. Substituting the given values, XL = 40 V / (2 A * 2 * π * 130 Hz) = 40 V / (260 A * π Hz) = 0.061 ohms.

Since XL = 2 * π * f * L, we can rearrange the equation and solve for L, L = XL / (2 * π * f).

Substituting the calculated XL and given f, L = 0.061 ohms / (2 * π * 130 Hz) = 0.000118 H (or 118 µH).