The instantaneous voltages at three terminals marked X, Y and Z are given by
$V_{x} = V_{0} sin ω t$
$V_{y} = V_{0} sin (ω t + \frac{2 π}{3}) a n d$
$V_{z} = V_{0} sin (ω t + \frac{4 π}{3})$
An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z.The reading (s) of the voltmeter will be

{V_{X Y}}^{r m s} = V_{0} \sqrt{\frac{3}{2}}

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{V_{Y Z}}^{r m s} = V_{0} \sqrt{\frac{1}{2}}

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{V_{Y Z}}^{r m s} = V_{0}

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Independent of the choice of the two terminals

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Solution

The correct options are
A ${V_{X Y}}^{r m s} = V_{0} \sqrt{\frac{3}{2}}$
D Independent of the choice of the two terminals
Potential difference
$V_{x y} = V_{x} - V_{y} V_{x y} = (V_{x y}) sin (ω t + θ_{1}) w h e r e (V_{x y})_{0} = \sqrt{V_{0}^{2} + V_{0}^{2} - 2 V_{0}^{2} cos \frac{2 π}{3}} = \sqrt{3} V_{0}$
and $(V_{x y})_{r m s} = \frac{(V_{x y})_{0}}{\sqrt{2}} = \sqrt{\frac{3}{2}} V_{0}$
(A) is correct
$V_{y z} = V_{y} - V_{z} V_{y z} = (V_{y z})_{0} sin (ω t + θ_{2}) w h e r e (V_{y z})_{0} = \sqrt{V_{0}^{2} + V_{0}^{2} - 2 V_{0}^{2} cos \frac{2 π}{3}} = \sqrt{3} V_{0}$
and $(V_{y z})_{r m s} = \frac{(V_{y z})_{0}}{\sqrt{2}} = \sqrt{\frac{3}{2}} V_{0}$
(D) is correct

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The instantaneous voltages at three terminals marked X, Y and Z are given by

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