Question

The instantaneous voltages at three terminals marked X,Y and Z are given by
$V_x=V_0\sin \omega t,$
$V_x=V_0\sin \sin (\omega t+\frac{2\pi }{3})$ and
$V_x=V_0\sin \sin (\omega t+\frac{4\pi }{3})$
An ideal voltmeter is configured to read rms value of the potential difference between its terminal. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be

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

Answer 1

The correct option is B $V_{XY}^{rms}=V_o\sqrt{\frac{3}{2}}$

$V_{XY}=V_X-V_Y$
$V_{XY}={\left(V_{XY}\right)}_0\sin (\omega t+Q_1)$
Where ${\left(V_{XY}\right)}_0=\sqrt{V_0^2+V_0^2-{2V}_0^2\cos \frac{2\pi }{3}}=\sqrt{3}{V}_0$
and $V_{XY}^{rms}=\frac{{\left(V_{XY}\right)}_0}{\sqrt{2}}=\sqrt{\frac{3}{2}}{V}_0$
$V_{YZ}=V_Y-V_Z$
$V_{YZ}={\left(V_{YZ}\right)}_0\sin (\omega t+Q_2)$
Where ${\left(V_{YZ}\right)}_0=\sqrt{V_0^2+V_0^2-{2V}_0^2\cos \frac{2\pi }{3}}=\sqrt{3}{V}_0$
and $V_{YZ}^{rms}=\frac{{\left(V_{YZ}\right)}_0}{\sqrt{2}}=\sqrt{\frac{3}{2}}{V}_0$