QJEEAdv17Ph4f49: JEE Advanced 2017 Physics - Class 12 JEEAdv17Ph4f49 Solution: The instantaneous voltages at three terminals marked and are…

The potential difference between point X and Y is given by

V_XY = V_X $-$ V_Y

= 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

V_YZ = V_Y $-$ V_Z

= 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

V_ZX = V_Z $-$ V_X

= 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 V^rms = 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.