In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ℰ^0 = 50 V and v = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential difference across the three elements is greater than the rms voltage of the source.

Sol. R = 300Ω, C = 20 μF = 20 * 10^-6 F L = 1Henry, E = 50 V V = 50/π Hz (a) I base 0 = E base 0/Z , Z = √R^2 + (X base c – X base L)^2 = √(300)^2 + (1/2πfC - 2πfL)^2 = √(300)^2 + (1/2π* 50/π *20 * 10^-6 - 2π * 50/π * 1)^2 = √(300)^2 +(10^4/20 - 100)^2 = 500 I base 0 E base 0/Z = 50/500 = 0.1 A (b) Potential across the capacitor = I base 0 × X base c = 0.1 × 500 = 50 V Potential difference across the resistor = I base 0 × R = 0.1 × 300 = 30 V Potential difference across the inductor = I base 0 × X base L = 0.1 × 100 = 10 V Rms. potential = 50 V Net sum of all potential drops = 50 V + 30 V + 10 V = 90 V Sum or potential drops > R.M.S potential applied.