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1.	Find the value of the source current from the following circuit.(a) 2.54A(b) 6.67A(c) 3.35A(d) 7.65AThe question was asked during a job interview.Query is from Kirchhoff’s Laws and Network Solution in chapter Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» The CORRECT OPTION is (a) 2.54A Explanation: I3 =(3+j0)A V2 =I3R=(3+j0)(8+j0)=(24+j0)V I2=V2/Xc=(j1.5) A I1 =I2 +I3 =(0+j1.5)+(3+j0)=(3+j1.5)A I1=(3^2+1.5^2)^1/2= 3.35A.

2.	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 current in the circuit.(a) 0.01 A(b) 0.2 A(c) 0.02 A(d) 0.002 AThe question was asked during an interview.The above asked question is from Kirchhoff’s Laws and Network Solution topic in chapter Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» The correct answer is (B) 0.2 A Explanation: Z^2 = R^2+(XL-XC)^2, 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/(2π5015*10^-6) =212.21 ohm. Z^2=50^2+(212.21-94.25)^2=15625 Z = 125 ohm. Current in the CIRCUIT i=V/Z=25/125=0.2 A.

3.	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 inductive reactance in the circuit.(a) 36 ohm(b) 95 ohm(c) 125 ohm(d) 140 ohmI got this question in an online interview.My question comes from Kirchhoff’s Laws and Network Solution topic in division Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» RIGHT OPTION is (B) 95 ohm Easy explanation: XL=2πfLf=50Hz and L=0.3H XL=2π(50)(0.3) = 94.25 ohm.

4.	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/(2π5015*10^-6) =212.21 ohm. tanϕ = \|(XL-XC)\|/R = (212.21-94.25)/50 = 2.3592 ϕ=67^0.

5.	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 impedance in the circuit.(a) 110 ohm(b) 100 ohm(c) 125 ohm(d) 140 ohmThis question was posed to me in an international level competition.The question is from Kirchhoff’s Laws and Network Solution topic in portion Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» The correct answer is (c) 125 OHM Explanation: Z^2 = R^2+(XL-XC)^2, 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/(2π5015*10^-6) =212.21 ohm. Z^2=50^2+(212.21-94.25)^2=15625 Z = 125 ohm.

6.	What value of direct current must flow through a resistor to produce the same heating power as an alternating current with a peak value of 3.5 A?(a) 1.5 A(b) 2.5 A(c) 3.5 A(d) 4.5 AThis question was addressed to me in semester exam.Enquiry is from Kirchhoff’s Laws and Network Solution topic in division Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» Correct choice is (B) 2.5 A To explain: Power in DC circuit = Power in ac circuit(GIVEN) IDC^2R = IRMS^2R IDC=IRMS IRMS=I0/√2 = 3.5/√2 = 2.5 A. IDC=2.5 A.

7.	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 capacitive reactance in the circuit.(a) 316 ohm(b) 195 ohm(c) 124 ohm(d) 212 ohmThe question was posed to me during an online interview.The above asked question is from Kirchhoff’s Laws and Network Solution topic in portion Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» CORRECT answer is (d) 212 ohm The best explanation: XC=1/(2πfC)f=50Hz and C=15 μF XC=1/(2π5015*10^-6) =212.21 ohm.

8.	Find the value of the source voltage from the following circuit.(a) 49.2V(b) 34.6V(c) 65.2V(d) 25.6VI got this question in an interview for internship.This interesting question is from Kirchhoff’s Laws and Network Solution in chapter Network Theorems Applied to AC Networks of Basic Electrical Engineering
Answer» The CORRECT choice is (a) 49.2V For explanation I WOULD say: I3 =(3+j0)A V2 =I3R=(3+j0)(8+j0)=(24+j0)V I2=V2/Xc=(0 + j1.5) A I1 =I2 +I3 =(0+j1.5)+(3+j0)=(3+j1.5)A I1=(3^2+1.5^2)^1/2= 3.35A. V1 =I1(R+jXL) =(15+j30)V E=V1 +V2 =(39+j30)V E=(39^2+30^2)^1/2= 49.2V.