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PhyMedAdv 2017 • Paper 2

Rms Potential Difference Across Terminals in Three Phase Supply

Consider three terminals designated as XX, YY, and ZZ with instantaneous voltages given by VX=V0sin⁡ωt,V_X = V_0 \sin \omega t, VY=V0sin⁡(ωt+2π3) andV_Y = V_0 \sin \left(\omega t + \frac{2\pi}{3}\right) \text{ and} VZ=V0sin⁡(ωt+4π3).V_Z = V_0 \sin \left(\omega t + \frac{4\pi}{3}\right). An ideal voltmeter measures the rmsrms value of potential difference between its input terminals. If it is connected across XX and YY and subsequently across YY and ZZ, the potential reading(s) of the voltmeter will be

Options

A

VXYrms=V032V_{XY}^{rms} = V_0 \sqrt{\frac{3}{2}}

Correct
B

VYZrms=V012V_{YZ}^{rms} = V_0 \sqrt{\frac{1}{2}}

C

VXYrms=V0V_{XY}^{rms} = V_0

D

independent of the choice of the two terminals

Correct

Topics & Concepts

Alternating CurrentAC Circuits

Step-by-Step Solution

To find the rmsrms (root mean square) value of the potential difference between any two terminals, we can calculate the instantaneous potential difference and find its peak value, or use the phasor representation of alternating voltages.

1. Instantaneous Potential Difference Calculation

The instantaneous potential difference between terminals XX and YY is given by: VXY=VX−VY=V0sin⁡ωt−V0sin⁡(ωt+2π3)V_{XY} = V_X - V_Y = V_0 \sin \omega t - V_0 \sin \left(\omega t + \frac{2\pi}{3}\right)

Using the trigonometric identity sin⁡A−sin⁡B=2sin⁡(A−B2)cos⁡(A+B2)\sin A - \sin B = 2 \sin\left(\frac{A - B}{2}\right) \cos\left(\frac{A + B}{2}\right): Let A=ωtA = \omega t and B=ωt+2π3B = \omega t + \frac{2\pi}{3}. A−B2=−π3\frac{A - B}{2} = -\frac{\pi}{3} A+B2=ωt+π3\frac{A + B}{2} = \omega t + \frac{\pi}{3}

Substituting these back into the expression: VXY=V0[2sin⁡(−π3)cos⁡(ωt+π3)]V_{XY} = V_0 \left[ 2 \sin\left(-\frac{\pi}{3}\right) \cos\left(\omega t + \frac{\pi}{3}\right) \right]

Since sin⁡(−π3)=−32\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}: VXY=−3V0cos⁡(ωt+π3)V_{XY} = -\sqrt{3} V_0 \cos\left(\omega t + \frac{\pi}{3}\right)

The peak value (amplitude) of the AC voltage VXYV_{XY} is: (VXY)0=3V0(V_{XY})_0 = \sqrt{3} V_0

The rmsrms value of a sinusoidal AC voltage with peak value (VXY)0(V_{XY})_0 is: VXYrms=(VXY)02=3V02=V032V_{XY}^{rms} = \frac{(V_{XY})_0}{\sqrt{2}} = \frac{\sqrt{3} V_0}{\sqrt{2}} = V_0 \sqrt{\frac{3}{2}}

2. Potential Difference Between Terminals YY and ZZ

Similarly, for terminals YY and ZZ: VYZ=VY−VZ=V0sin⁡(ωt+2π3)−V0sin⁡(ωt+4π3)V_{YZ} = V_Y - V_Z = V_0 \sin \left(\omega t + \frac{2\pi}{3}\right) - V_0 \sin \left(\omega t + \frac{4\pi}{3}\right)

Applying the same trigonometric transformation: VYZ=V0[2sin⁡(−π3)cos⁡(ωt+π)]=3V0cos⁡(ωt)V_{YZ} = V_0 \left[ 2 \sin\left(-\frac{\pi}{3}\right) \cos(\omega t + \pi) \right] = \sqrt{3} V_0 \cos(\omega t)

The peak value of VYZV_{YZ} is also (VYZ)0=3V0(V_{YZ})_0 = \sqrt{3} V_0. Therefore, its rmsrms value is: VYZrms=3V02=V032V_{YZ}^{rms} = \frac{\sqrt{3} V_0}{\sqrt{2}} = V_0 \sqrt{\frac{3}{2}}

3. Phasor Representation Method

The three voltages form a balanced three-phase system with equal amplitudes V0V_0 and phase difference of 120∘120^\circ (2π3\frac{2\pi}{3} rad) between adjacent terminals.

The peak potential difference between any two terminals is the magnitude of the vector difference between their respective phasors: Vpeak=V02+V02−2V02cos⁡120∘=V02+V02+V02=3V0V_{\text{peak}} = \sqrt{V_0^2 + V_0^2 - 2 V_0^2 \cos 120^\circ} = \sqrt{V_0^2 + V_0^2 + V_0^2} = \sqrt{3} V_0

Hence, the rmsrms voltage measured across any pair of terminals (XYXY, YZYZ, or ZXZX) is: Vrms=Vpeak2=V032V^{rms} = \frac{V_{\text{peak}}}{\sqrt{2}} = V_0 \sqrt{\frac{3}{2}}

This result shows that:

  1. VXYrms=V032V_{XY}^{rms} = V_0 \sqrt{\frac{3}{2}} (Option A is correct).
  2. The rmsrms potential difference is the same for any pair of terminals, meaning it is independent of the choice of the two terminals (Option D is correct).

Correct Options: A and D

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