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Phasor Diagrams in AC Circuits

The instantan...

Question

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

$V_{X} = V_{0} sin ω t$

$V_{Y} = V_{0} sin (ω t + \frac{2 π}{3})$ and

$V_{Z} = V_{0} sin (ω_{0} + \frac{4 π}{3})$

An ideal voltmeter is configured to read rms of the value of potential difference between its terminals. It is connected between points 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_{o} \sqrt{\frac{3}{2}}

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(V_{Y Z})^{r m s} = V_{o} \sqrt{\frac{1}{2}}

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$(V_{X Y})^{r m s} = V_{o}$

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Independent of the choice of the two terminals

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Solution

The correct option is D
Independent of the choice of the two terminals

The potential diffrence between $X$ and $Y$ is

$V_{X Y} = V_{X} - V_{Y}$
$V_{X Y} = (V_{X Y})_{0} s i n (ω t + θ_{1})$
Where $(V_{X Y})_{0} = \sqrt{V_{0}^{2} + V_{0}^{2} - 2 V_{0}^{2} c o s \frac{2 π}{3}} = \sqrt{3} V_{0}$
and $(V_{X Y})_{r m s} = \frac{(V_{X Y})_{0}}{\sqrt{2}} \sqrt{\frac{3}{2}} V_{0}$
$∴$ option A is the correct option ,
now the potential diffrence between $Y$ and $Z$ is

$V_{Y Z} = V_{Y} - V_{Z}$
$V_{Y Z} = (V_{Y Z})_{0} s i n (ω t + θ_{2})$
Where $(V_{Y Z})_{0} = \sqrt{V_{0}^{2} + V_{0}^{2} - 2 V_{0}^{2} c o s \frac{2 π}{3}} = \sqrt{3} V_{0}$
and $(V_{Y Z})_{r m s} = \frac{(V_{Y Z})_{0}}{\sqrt{2}} \sqrt{\frac{3}{2}} V_{0}$
$∴$ option D is the correct answer

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Similar questions

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

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