JEE Advance - Physics (2017 - Paper 2 Offline - No. 3)
$${V_x} = {V_0}\,\sin \,\omega t,$$
$${V_Y} = {V_0}\,\sin $$ $$\left( {\omega t + {{2\pi } \over 3}} \right)$$
and $$Vz = {V_0}\sin \left( {\omega t + {{4\pi } \over 3}} \right)$$
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
Explanation
The potential difference between point X and Y is given by
VXY = VX $$-$$ VY
= V0 sin($$\omega$$t) $$-$$ V0 sin($$\omega$$t + $${{2\pi } \over 3}$$)
= 2V0 cos($$\omega$$t + $${{\pi } \over 3}$$) sin ($$-$$$${{\pi } \over 3}$$)
= $$-$$$$\sqrt 3 $$V0 cos($$\omega$$t + $${\pi \over 3}$$) = $$\sqrt 3 $$V0 sin ($$\omega$$t $$-$$ $${\pi \over 6}$$).
Similarly, the potential difference between point Y and Z is
VYZ = VY $$-$$ VZ
= V0 sin($$\omega$$t + $${2\pi \over 3}$$) $$-$$ V0 sin($$\omega$$t + $${4\pi \over 3}$$)
= 2V0 cos($$\omega$$t + $$\pi$$) sin ($$-$$$${\pi \over 3}$$)
= $$\sqrt3$$V0 cos($$\omega$$t) = $$\sqrt3$$V0 sin($$\omega$$t + $${\pi \over 2}$$),
and the potential difference between point Z and X is
VZX = VZ $$-$$ VX
= V0 sin($$\omega$$t + $${4\pi \over 3}$$) $$-$$ V0 sin($$\omega$$t)
= $$\sqrt3$$V0 cos($$\omega$$ + $${2\pi \over 3}$$) = $$\sqrt3$$V0 sin($$\omega$$t + $${7\pi \over 6}$$).
The rms value of the potential V = V0 sin($$\omega$$t + $$\phi$$) is given by Vrms = V0 / $$\sqrt2$$. Thus, rms values of the potentials are $$V_{XY}^{rms} = V_{YZ}^{rms} = V_{ZX}^{rms} = {V_0}\sqrt {3/2} $$. Hence, the reading of the voltmeter is independent of the two terminal i.e., reading is same whether it is connected across X-Y or Y-Z or Z-X.
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