Question

Derive equations 1 and 2, thank you!

When you have a RC series circuit in an AC source of angular frequency ω (ω = 2πf), you can calculate the capacitance (C) from the equation: max, where VmarR is the amplitude of the voltage across the resistor and Vmax.c is the amplitude of the voltage across the capacitor. [You have to derive this equation in your report.] Similarly, when you have a RL series circuit in an AC source of angular frequency o 2τή, you can calculate the inductance (L) from the equation: (c) max, R where Vmax.R is the amplitude of the voltage across the resistor, Vmax L is the amplitude of the voltage across the inductor, and Ri is the resistance of the inductor and R is the external resistance connected in series to the inductor (that has resistance Ri). [You have to derive this equation as well in your report.]

Answer 1

Data given: - An RC series circuit in an AC source of angular frequency omega (omega = 2 pi t) calculation of capacitance (c) from the equation. c = V_max middot R/V_max middot C = 1/omega R. where V_max R is the amplitude of the voltage across the Resistance. V_max c is the amplitude of the voltage across the capacitor. In R-C circuit voltage amplitude across Resistance = V_m R voltage amplitude across capacitor = V_max C. \r\n As R and C are in series current passing through R and C is done. I_R = I V_max R/R = V_max_1 C/X_c} where X_c is resistance of capacitor V_max_1 R/R = V_max C/1 middot omega C x_c = 1/omega c c = (V_max R/V_max C) 1/omega R PL-source circuit in an AC source of angular frequency omega (omega = 2 pi f) (omega c)^2 = (R V_max middot L/V_max middot R)^2 - RL^2 \r\n voltage amplitude across capacitor inductor = V_max voltage amplitude