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Alternating Current

Section: Physics

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Question : 5 of 13

Marks: +1, -0

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

V_X = V_0 \sin \omega t

V_Y = V_0 \sin\left(\omega t + \frac{2\pi}{3}\right)

V_Z = V_0 \sin\left(\omega t + \frac{4\pi}{3}\right)

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

[JEE Adv 2017 P2]

$V^{rms}_{YZ} = V_0 \sqrt{\frac{1}{2}}$
$V^{rms}_{YZ} = V_0 \sqrt{\frac{3}{2}}$
$V^{rms}_{YZ} = V_0$
independent of the choice of the two terminals

Solution:

If connected across X and Y :

V_{XY} = V_X - V_Y

= V_0 \sin\left(\omega t + \frac{2\pi}{3}\right)

= V_0 [ \sin \omega t - \sin \omega t \cos \omega t \frac{2\pi}{3} -

\cos \omega t \sin \frac{2\pi}{3} ]

= V_0 \left[ \sin \omega t + \frac{1}{2} \sin \omega t - \frac{\sqrt{3}}{2} \cos \omega t \right]

= \sqrt{3} V_0 \left[ \frac{\sqrt{3}}{2} \sin \omega t - \frac{1}{2} \cos \omega t \right]

= \sqrt{3} V_0 \sin\left(\omega t - \frac{\pi}{3}\right)

\therefore V^{rms}_{XY}

= \frac{\sqrt{3} V_0}{\sqrt{2}}

= \sqrt{\frac{3}{2}} V_0

If connected across X and Z.

V_{XZ} = V_X - V_Z

= V_0 \sin \omega t - V_0 \sin\left(\omega t + \frac{4\pi}{3}\right)

= \sqrt{3} V_0 \sin\left(\omega + \frac{\pi}{6}\right)

If connected across Y and Z

V_{YZ} = V_Y - V_Z

= V_0 \sin\left(\omega t + \frac{2\pi}{3}\right)

- V_0 \sin\left(\omega t + \frac{4\pi}{3}\right)

= \sqrt{3} V_0 \cos \omega t

\therefore V_{rms}

is same if the voltmeter is connected across an

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