Locus Diagrams MCQ Quiz - Objective Question with Answer for Locus Diagrams - Download Free PDF

Last updated on Apr 14, 2025

Latest Locus Diagrams MCQ Objective Questions

Locus Diagrams Question 1:

A 1 - ϕ circuit with a supply voltage 'V' consists of resistance 'R’ and reactance 'X' in series. The locus of current is a circle with a radius of when ‘R’ is varied.

  1. V/X
  2. V/2X
  3. V/πX
  4. V/2πX

Answer (Detailed Solution Below)

Option 2 : V/2X

Locus Diagrams Question 1 Detailed Solution

Locus diagram:

The locus diagram is useful for the analysis and designing of the circuit.

Ex: Filters

Current locus diagram:

The single-phase circuit with series R-L is shown below.

F1 Jai 26.10.20 Pallavi D23

R= Resistance

X= Inductive reactance.

the effect of variation in R is shown below

R

Z = R + j X

I = (V/Z)

Θ = 

tan-1 (X/R)

0

X

V/X

900

Increases

increases

Decreases

decreases

0

00

The current locus diagram for the above circuit is shown below.

F1 Jai 26.10.20 Pallavi D24

From the diagram,

Diameter = (V/X)

Radius = (V/2X)

Centre = (0, (V/2X))

Important Points

  • When the power factor angle is variable, the shape of the current locus is a semi-circle.
  • When the power factor angle is constant, the shape of the current locus is a Straight line.

Top Locus Diagrams MCQ Objective Questions

Locus Diagrams Question 2:

A 1 - ϕ circuit with a supply voltage 'V' consists of resistance 'R’ and reactance 'X' in series. The locus of current is a circle with a radius of when ‘R’ is varied.

  1. V/X
  2. V/2X
  3. V/πX
  4. V/2πX

Answer (Detailed Solution Below)

Option 2 : V/2X

Locus Diagrams Question 2 Detailed Solution

Locus diagram:

The locus diagram is useful for the analysis and designing of the circuit.

Ex: Filters

Current locus diagram:

The single-phase circuit with series R-L is shown below.

F1 Jai 26.10.20 Pallavi D23

R= Resistance

X= Inductive reactance.

the effect of variation in R is shown below

R

Z = R + j X

I = (V/Z)

Θ = 

tan-1 (X/R)

0

X

V/X

900

Increases

increases

Decreases

decreases

0

00

The current locus diagram for the above circuit is shown below.

F1 Jai 26.10.20 Pallavi D24

From the diagram,

Diameter = (V/X)

Radius = (V/2X)

Centre = (0, (V/2X))

Important Points

  • When the power factor angle is variable, the shape of the current locus is a semi-circle.
  • When the power factor angle is constant, the shape of the current locus is a Straight line.

Locus Diagrams Question 3:

The admittance locus of a two-branch parallel circuit shown in fig.1 is given in fig.2. Assuming ω = √3 rad / sec, find the values of R1, R2 and C2

quesOptionImage334

  1. 2Ω, 2Ω,  23
  2. 0.5Ω, 2Ω, 32F
  3. 0.5Ω, 0.5Ω, 23F
  4. 2Ω, 5Ω, 32F

Answer (Detailed Solution Below)

Option 3 : 0.5Ω, 0.5Ω, 23F

Locus Diagrams Question 3 Detailed Solution

Concept:

For the given locus diagram C1 is variable and R1 is constant so, its admittance locus will form a semi-circle with diameter 1/R1

R2, C2 are constant so its admittance locus will form a straight line passing through the origin

So, total admittance is the sum of individual admittances

And the total admittance locus is as shown below:

F1 Jai.P 9-11-20 Savita D7

Given admittance locus:

F1 Jai.P 9-11-20 Savita D8

Impedance Z1 = R1 + jXc1

Admittance Y1=R1+jXc1R12+Xc12=R1R12+Xc12+j(Xc1R12+Xc12)

Compare the above equations with Y1 = G1 + jB1

G1=R1R12+Xc12;B1=Xc1R12+Xc12

Similarly,

Impedance Z2 = R2 + jXc2

Admittance Y2=R2+jXc2R22+Xc22=R2R22+Xc22+j(Xc2R22+Xc22)

Compare the above equations with Y2 = G2 + jB2

G2=R2R22+Xc22;B2=Xc2R22+Xc22

Where,

R1, R2 are resistors of impedances Z1, Z2 respectively

Xc1, Xc2 are capacitive reactance(Xc = 1/ωC) of impedances Z1, Z2 respectively

Y1, Y2 are admittances of impedances Z1 and Z2 respectively

G1 and G2 are conductances of admittances Y1, Y2 respectively

B1, B2 are susceptances of admittances Y1, Y2 respectively

Calculation:

From the graph:

The diameter of the semi-circle 1/R1 = 2s

R1 = 0.5Ω

tanθ = B2/G2

= 0.866/0.5

= 1.732 = √3

B2/G2 = Xc2/R2 = 1/ωC2R2

= 1/ωC2R2 = √3 (given ω = √3)

= R2C2 = 1/3

By verifying options, In option (3) R2C2 = 0.5×2/3 = 1/3

So, R1 = R2 = 0.5Ω, C2 = 2/3 F

Locus Diagrams Question 4:

A series R-C circuit has constant value of R and varying value of C. The current locus with constant applied voltage V at constant frequency, as C is varies, is a semi circle with centre at

  1. (VR,0)
  2. (V2R,0)
  3. (V4R,0)
  4. (V2Xc,0)

Answer (Detailed Solution Below)

Option 2 : (V2R,0)

Locus Diagrams Question 4 Detailed Solution

In a series RC circuit, let

f = constant frequency

R = constant resistance

C = varying capacitor

Vs = constant applied voltage

Let the capacitor is initially fully discharged.

When the voltage is applied to the RC circuit, the capacitor draws a charging current and charges up and when the voltage is reduced then the capacitor will get discharged.

Now, z=R2+XC2,XC=1ωC 

And IC=Vz

F1 Uday Madhu 25.07.20 D6

Ic=VR2+XC2

Now, when C increases then XC is decreased, and hence the current I also increases.

Now divide the current in x and y component

And IC2=Ix2+Iy2

F1 Uday Madhu 25.07.20 D7

Ix2+Iy2=V2R2+XC2         ---(1)

When XC = 0, then IC is maximum and θ = 0, cos θ = 1 (Power factor)

If XC is increasing, then the current reduces and finally, θ is leading (the voltage) and cosθ=Rz

Now, Ix=Iccosθ=VZRZ=VRZ2         ---(2)

And Iy=Icsinθ=VZXcZ=VXCZ2      ---(3)

Now, V2Z2 in terms of R, because R is constant,

 put in equation (1)

Ix2+Iy2=IxVR 

Ix2IxVR+(V2R)2(V2R)2+Iy2=0 

(IxV2R)2+Iy2=(V2R)2 

This equation represents a circle whose radius is V2R and centre is (V2R,0)

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