# MCQs on Components Interconnection and Impedance Matching

1 - Question

What is the equivalent capacitance when two capacitors, one of 20 microfarads and another of 10 microfarads are connected in series?
Explanation: The equivalent capacitance when two capacitors, one of 20 microfarads and another of 10 microfarads are connected in series is 15 microfarads. When capacitors are placed is series to each other then the equivalent capacitance is given by 1/Ct= (1/C1)+(1/C2)+….(1/Cn), where Ct is the total capacitance. C1,C2….Cn are total no. of capacitors in series. Therefore (1/Ct)=(1/20)+(1/10) Ct=15 microfarads.

2 - Question

What is the effect on phase shift due to a capacitor in AC circuits?
a) Voltage leads the current by 180 degree
b) Voltage leads the current by 90 degree
c) Current leads the voltage by 90 degree
d) Current leads the voltage by 180 degree
Explanation: The phase shift due to a capacitor in AC circuits is 90 degrees. In this the Current leads the voltage by 90 degree. While in case of an inductor the phase difference remain same but the current lags the voltage by 90 degree.

3 - Question

What is the source voltage applied across a loop containing a resistor of 1000 ohm and capacitor of 15 microfarads in series when a current of 0.02 ampere flows across the circuit? Given that the voltage drop across capacitor is 5 Volts?
a) 25 Volts
b) 24 Volts
c) 26 Volts
d) 28 Volts
Explanation: Given : Voltage drop across capacitor = 5 Volts Current in loop = 0.02 ampere Resistance = 1000 ohm Capacitance=15 microfarads Drop across resistor=1000 *0.02 Drop across resistor=20 Volts Source voltage (Total Voltage) = Drop across resistor + Drop across resistor Source voltage = 10 Volts + 5 Volts Source voltage=25 Volts. advertisement

4 - Question

What is the equivalent capacitance when two capacitors each of 20 microfarads are connected in parallel?
Explanation: The equivalent capacitance when two capacitors each of 20 microfarads are connected in parallel is 40 microfarads. When capacitors are placed parallel to each other, then their value of capacitances simply gets added up.

5 - Question

What is the reactance due to a capacitor of 0.05 farads when an alternating current of frequency 50 Hertz passes through it?
a) 0.063 ohm
b) 0.63 ohm
c) 0.0063 ohm
d) 0.0703 ohm
Explanation: Given: Frequency of alternating current(f) = 50 Hertz Capacitance(C) = 0.05 farads Reactance(Xc)=1/2πfC Reactance(Xc)=1/(2*3.14*50 *0.05) Reactance(Xc)=0.063 ohm.

6 - Question

What is the reactance due to an inductor of 0.05 henry when an alternating current of frequency 50 Hertz passes through it?
a) 16.7 ohm
b) 15.7 ohm
c) 17.7 ohm
d) 14.7 ohm
Explanation: Given: Frequency of alternating current(f) = 50 Hertz Inductance(L) = 0.05 henry Reactance(Xl)=2πfL Reactance(Xl)=2*3.14*50*0.05 Reactance(Xl)=15.7 ohm.

7 - Question

What is the equivalent Inductance when two Inductors one of 20 henry and another of 15 henry are connected in series?
a) 20 henry
b) 35 henry
c) 40 henry
d) 10 henry
Explanation: The equivalent Inductance when two Inductors, one of 20 henry and another of 15 henry are connected in series is 35 henry. When Inductors are placed in series to each other, then their value of Inductance simply gets added up.

8 - Question

What is the equivalent Inductance when two Inductors, one of 20 henry and another of 10 henry are connected in parallel?
a) 20 henry
b) 30 henry
c) 15 henry
d) 10 henry
Explanation: The equivalent Inductance when two Inductors, one of 20 henry and another of 10 henry are connected in parallel is 15 henry. When Inductors are placed is parallel to each other then the equivalent Inductance is given by 1/Lt=(1/L1)+(1/L2)+….(1/Ln), where Lt is the total Inductance. L1,L2….Ln are total no. of inductors in series. Therefore (1/Lt)=(1/20)+(1/10) Lt=15 henry.

9 - Question

What is the equivalent capacitance when two capacitors each of 20 microfarads are connected in series?
Explanation: The equivalent capacitance when two capacitors each of 20 microfarads are connected in series is 10 microfarads. When capacitors are placed is series to each other then the equivalent capacitance is given by 1/Ct=(1/C1)+(1/C2)+….(1/Cn), where Ct is the total capacitance. C1,C2….Cn are total no. of capacitors in series. Therefore (1/Ct)=(1/20)+(1/20) Ct=10 microfarads.

10 - Question

What is the source voltage applied across a loop containing a resistor of 100 ohm and capacitor of 15 microfarads in series when a current of 0.05 ampere flows across the circuit? Given that the voltage drop across capacitor is 2 Volts?
a) 7 Volts
b) 4 Volts
c) 6 Volts
d) 8 Volts
Explanation: Given: Voltage drop across capacitor = 2 Volts Current in loop = 0.05 ampere Resistance = 100 ohm Capacitance=15 microfarads Drop across resistor=100*0.05 Drop across resistor=5 Volts Source voltage (Total Voltage) = Drop across resistor + Drop across resistor Source voltage = 5 Volts + 2 Volts Source voltage=7 Volts.

11 - Question

What is the reactance due to an inductor of 0.01 henry when an alternating current of frequency 40 Hertz passes through it?
a) 3.512 ohm
b) 15.7 ohm
c) 2.512 ohm
d) 14.7 ohm
Explanation: Given: Frequency of alternating current(f) = 40 Hertz Inductance(L) = 0.01 henry Reactance(Xl)=2πfL Reactance(Xl)=2*3.14*40*0.01 Reactance(Xl)=2.512 ohm.

12 - Question

What is the reactance due to a capacitor of 0.01 farads when an alternating current of frequency 40 Hertz passes through it?
a) 0.063 ohm
b) 0.398 ohm
c) 0.0063 ohm
d) 0.0703 ohm
Explanation: Given: Frequency of alternating current(f) = 40 Hertz Capacitance(C) = 0.01 farads Reactance(Xc)=1/2πfC Reactance(Xc)=1/(2*3.14*40 *0.01) Reactance(Xc)=0.398 ohm.

13 - Question

The impedance matching is done to match the internal impedance of two different sources.
a) True
b) False
Explanation: The impedance matching is not done to match the internal impedance of two different sources. Rather this technique is used to match the internal source resistance and the driven load resistance. It maximizes the power transfer.

14 - Question

What is the source voltage applied across a loop containing a resistor of 500 ohm and capacitor of 10 microfarads in series when a current of 0.02 ampere flows across the circuit? Given that the voltage drop across capacitor is 2 Volts.
a) 12 Volts
b) 4 Volts
c) 6 Volts
d) 8 Volts
Explanation: Given : Voltage drop across capacitor = 2 Volts Current in loop = 0.02 ampere Resistance = 500 ohm Capacitance=10 microfarads Drop across resistor=500*0.02 Drop across resistor=10 Volts Source voltage (Total Voltage) = Drop across resistor + Drop across resistor Source voltage = 10 Volts + 2 Volts Source voltage=12 Volts.

15 - Question

What is the effect on phase shift due to an inductor in AC circuits?
a) Voltage leads the current by 180 degree
b) Voltage leads the current by 90 degree
c) Current leads the voltage by 90 degree
d) Current leads the voltage by 180 degree