The instantaneous voltages at three terminals marked X, Y and Z are given by
`V_(X)=V_(0)sin omegat, V_(Y)=V_(0)sin (omegat=(2pi)/(3))and V_(z)=V_(0)sin (omegat=(4pi)/(3))`
An ideal voltmeter is configured to read runs value of the potential difference between its terminals. is connected between points X and Y and then between Y and Z. The reading the voltmeter will be

Options

A. `V_(X)=V_(0) sin omegat, V_(gamma)=V_(0)sin (omegat+(2pi)/(3))and V_(Z)=V_(0)sin (omegat+(4pi)/(3))`
B. `V_(XY)^(rms)=V_(0)sqrt((3)/(2))`
C. `V_(YZ)^(rms)=V_(0)sqrt((1)/(2))`
D. indpependent of the choice of the two terminals

Answer

To solve the problem, we need to calculate the potential difference readings of an ideal voltmeter connected between terminals X and Y, and then between terminals Y and Z. The instantaneous voltages at the terminals are given as follows:

1. \( V_X = V_0 \sin(\omega t) \)
2. \( V_Y = V_0 \sin\left(\omega t + \frac{2\pi}{3}\right) \)
3. \( V_Z = V_0 \sin\left(\omega t + \frac{4\pi}{3}\right) \)

### Step 1: Calculate the potential difference \( V_{XY} \)

The potential difference between terminals X and Y is given by:

\[ V_{XY} = V_X - V_Y \]

Substituting the values of \( V_X \) and \( V_Y \):

\[ V_{XY} = V_0 \sin(\omega t) - V_0 \sin\left(\omega t + \frac{2\pi}{3}\right) \]

Factoring out \( V_0 \):

\[ V_{XY} = V_0 \left( \sin(\omega t) - \sin\left(\omega t + \frac{2\pi}{3}\right) \right) \]

### Step 2: Apply the sine subtraction formula

Using the sine subtraction formula \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \):

Let \( A = \omega t \) and \( B = \omega t + \frac{2\pi}{3} \):

\[ V_{XY} = V_0 \cdot 2 \cos\left(\frac{(\omega t) + (\omega t + \frac{2\pi}{3})}{2}\right) \sin\left(\frac{(\omega t) - (\omega t + \frac{2\pi}{3})}{2}\right) \]

Calculating the averages:

\[ \frac{A+B}{2} = \omega t + \frac{\pi}{3} \]
\[ \frac{A-B}{2} = -\frac{\pi}{3} \]

Thus,

\[ V_{XY} = V_0 \cdot 2 \cos\left(\omega t + \frac{\pi}{3}\right) \sin\left(-\frac{\pi}{3}\right) \]

Since \( \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \):

\[ V_{XY} = -V_0 \cdot \sqrt{3} \cos\left(\omega t + \frac{\pi}{3}\right) \]

### Step 3: Calculate RMS value for \( V_{XY} \)

The peak value of \( V_{XY} \) is \( \sqrt{3} V_0 \). The RMS value is given by:

\[ V_{RMS, XY} = \frac{V_{peak}}{\sqrt{2}} = \frac{\sqrt{3} V_0}{\sqrt{2}} = \frac{V_0 \sqrt{3}}{\sqrt{2}} \]

### Step 4: Calculate the potential difference \( V_{YZ} \)

Now, we calculate the potential difference between terminals Y and Z:

\[ V_{YZ} = V_Y - V_Z \]

Substituting the values of \( V_Y \) and \( V_Z \):

\[ V_{YZ} = V_0 \sin\left(\omega t + \frac{2\pi}{3}\right) - V_0 \sin\left(\omega t + \frac{4\pi}{3}\right) \]

Factoring out \( V_0 \):

\[ V_{YZ} = V_0 \left( \sin\left(\omega t + \frac{2\pi}{3}\right) - \sin\left(\omega t + \frac{4\pi}{3}\right) \right) \]

### Step 5: Apply the sine subtraction formula again

Using the same sine subtraction formula:

Let \( A = \omega t + \frac{2\pi}{3} \) and \( B = \omega t + \frac{4\pi}{3} \):

\[ V_{YZ} = V_0 \cdot 2 \cos\left(\frac{(\omega t + \frac{2\pi}{3}) + (\omega t + \frac{4\pi}{3})}{2}\right) \sin\left(\frac{(\omega t + \frac{2\pi}{3}) - (\omega t + \frac{4\pi}{3})}{2}\right) \]

Calculating the averages:

\[ \frac{A+B}{2} = \omega t + \pi \]
\[ \frac{A-B}{2} = -\frac{\pi}{3} \]

Thus,

\[ V_{YZ} = V_0 \cdot 2 \cos\left(\omega t + \pi\right) \sin\left(-\frac{\pi}{3}\right) \]

Since \( \cos(\omega t + \pi) = -\cos(\omega t) \) and \( \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \):

\[ V_{YZ} = V_0 \cdot 2 \left(-\cos(\omega t)\right) \left(-\frac{\sqrt{3}}{2}\right) = V_0 \sqrt{3} \cos(\omega t) \]

### Step 6: Calculate RMS value for \( V_{YZ} \)

The peak value of \( V_{YZ} \) is \( \sqrt{3} V_0 \). The RMS value is given by:

\[ V_{RMS, YZ} = \frac{V_{peak}}{\sqrt{2}} = \frac{\sqrt{3} V_0}{\sqrt{2}} = \frac{V_0 \sqrt{3}}{\sqrt{2}} \]

### Conclusion

The readings of the voltmeter when connected between terminals X and Y, and then Y and Z, are both:

\[ V_{RMS, XY} = \frac{V_0 \sqrt{3}}{\sqrt{2}}, \quad V_{RMS, YZ} = \frac{V_0 \sqrt{3}}{\sqrt{2}} \]