The instantaneous voltage at three terminals marked x,y and z are given by . . .

Question : The instantaneous voltage at three terminals marked x,y and z are given by 

V_x=V₀sinωt

V_y=V₀sin(ωt+2π/3) and 

V_z=V₀sin(ωt+4π/3)

An idea voltmeter is configured to read rms value of the potential difference between its terminals. If is connected between point X and Y then between Y and Z. The reading(s) of the voltmeter will be

a) Independent of choice of the two terminals.

b) V_xy ^rms = V₀

c) V_yz ^rms = V₀√(1/2)

d) V_xy ^rms = V₀√(3/2)

Doubt by Mohini

Solution : 

V_x=V₀sinωt

V_y=V₀sin(ωt+2π/3) and 

V_z=V₀sin(ωt+4π/3)

First of all we are calculating the potential difference when the potentiometer is connected between X and Y

Phase Angle (Φ) between V_x and V_y

Φ = 2π/3-0
Φ = 2π/3
Now,
(V_o)_xy = √[(V_ox)²+(V_oy)²+2V_oxV_oycosΦ]

V_xy = √[V_o²+V_o²+2V_oV_ocos2π/3]
V_xy = √[2V_o²+2V_o²(-1/2)]
V_xy = √[2V_o²-V_o²]
V_xy = √[V_o²]
V_xy = V_o

V_xy ^rms = V_o/√2

V_xy ^rms = V_o/√(1/2)

Similarly 

Phase Angle (Φ) between V_y and V_z

Φ = 4π/3-2π/3
Φ = 2π/3
Now,
(V_o)_yz = √[V_oy²+V_oz²+2V_oyV_ozcosΦ]

V_yz = √[Vo²+V_o²+2V_oV_ocos2π/3]
V_yz = √[2V_o²+2V₀²(-1/2)]
V_yz = √[2Vo²-Vo²]
V_yz = √[Vo²]
V_yz = V_o

V_yz ^rms = V_o/√2

V_yz ^rms = V_o/√(1/2)

∵ V_xy ^rms = V_yz ^rms = V_o/√(1/2) 

Hence, a) and c) would be the correct option.