Question

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

$V_X=V_0\sin \omega t$

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

$V_Z=V_0\sin \left({\omega }_0+\frac{4\pi }{3}\right)$

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

• A. $(V_{XY}{)}^{rms}=V_o\sqrt{\frac{3}{2}}$
• B. $(V_{YZ}{)}^{rms}=V_o\sqrt{\frac{1}{2}}$
• C. $(V_{XY}{)}^{rms}=V_o$
• D. Independent of the choice of the two terminals

Answer 1

Answer: (A), (D)

Independent of the choice of the two terminals

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

$V_{XY}=V_X-V_Y$
$V_{XY}=\left(V_{XY}{)}_{0}sin\right(\omega t+{\theta }_1)$
Where $(V_{XY}{)}_0=\sqrt{V_0^2+V_0^2-2V_0^{2}cos\frac{2\pi }{3}}=\sqrt{3}{V}_0$
and $(V_{XY}{)}_{rms}=\frac{(V_{XY}{)}_0}{\sqrt{2}}\sqrt{\frac{3}{2}}{V}_0$
$\therefore$ option A is the correct option ,
now the potential diffrence between $Y$ and $Z$ is

$V_{YZ}=V_Y-V_Z$
$V_{YZ}=\left(V_{YZ}{)}_{0}sin\right(\omega t+{\theta }_2)$
Where $(V_{YZ}{)}_0=\sqrt{V_0^2+V_0^2-2V_0^{2}cos\frac{2\pi }{3}}=\sqrt{3}{V}_0$
and $(V_{YZ}{)}_{rms}=\frac{(V_{YZ}{)}_0}{\sqrt{2}}\sqrt{\frac{3}{2}}{V}_0$
$\therefore$ option D is the correct answer