Question

of 20 Ω, and a capacitance of 100 μF are connected in (Ctc +vZp m - YTZ )series across 220V, 50 Hz supply​

Answer 1

Explanation:

XC = 0.2 × 314 = 63Ω , C = 10μF = 100 × 10−6 = 10−4 farad XC = 1/ωC = 1/(314 x 10-4) = 32Ω, X = 63 − 32 = 31 Ω (inductive) (a) Z = √(202 + 31)2 = 37 Ω (b) I = 220/37 = 6 A (approx) (c) VR = I × R = 6 × 20 = 120 V; VL = 6 × 63 = 278 V, VC = 6 × 32 = 192 V (d) Power in VA = 6 × 220 = 1320 Power in watts = 6 × 220 × 0.54 = 713 W (e) p.f. = cos φ = R/Z = 20/37 = 0.54; φ = cos−1(0.54) = 57°18′Read more on Sarthaks.com - https://www.sarthaks.com/493275/resistance-of-20-an-inductance-of-and-capacitance-of-100-are-connected-in-series-across-220

Answer 2

Answer:

answer is

Explanation:

XC = 0.2 × 314 = 63Ω , C = 10μF = 100 × 10−6 = 10−4 farad XC = 1/ωC = 1/(314 x 10-4) = 32Ω, X = 63 − 32 = 31 Ω (inductive) (a) Z = √(202 + 31)2 = 37 Ω (b) I = 220/37 = 6 A (approx) (c) VR = I × R = 6 × 20 = 120 V; VL = 6 × 63 = 278 V, VC = 6 × 32 = 192 V (d) Power in VA = 6 × 220 = 1320 Power in watts = 6 × 220 × 0.54 = 713 W (e) p.f. = cos φ = R/Z = 20/37 = 0.54; φ = cos−1(0.54) = 57°18