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In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the phase difference between current and voltage.(a) 670(b) 540(c) 470(d) 770The question was posed to me in an international level competition.This interesting question is from Kirchhoff’s Laws and Network Solution in chapter Network Theorems Applied to AC Networks of Basic Electrical Engineering

Answer»

Right CHOICE is (a) 670

To explain: XL=2πfL,XC=1/(2πfC)f=50Hz and L=0.3H and C=15 μF

XL=2π(50)(0.3) = 94.25 ohm.

XC=1/(2π*50*15*10^-6) =212.21 ohm.

tanϕ = |(XL-XC)|/R = (212.21-94.25)/50 = 2.3592

ϕ=67^0.



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