The instantaneous voltages at three terminals marked X, Y and Z are given by V_X = V_0 sin(wt), V_Y

Q: Why do we use phasors instead of trigonometric identities?

Using trigonometric identities like $\sin A - \sin B$ works perfectly fine, but it can get algebraically messy. Phasors turn a trigonometry problem into a simple geometry problem, making it much easier to visualize and solve.

Q: Why is the angle between $V_X$ and $-V_Y$ $60^\circ$?

$V_X$ is at $0^\circ$ and $V_Y$ is at $120^\circ$. The negative of $V_Y$ points in the opposite direction, which is $120^\circ - 180^\circ = -60^\circ$. The angle between $0^\circ$ and $-60^\circ$ is exactly $60^\circ$.

Q: Why does the voltmeter read the RMS value?

In AC circuits, voltage and current are constantly changing. The RMS (Root Mean Square) value is the equivalent DC value that would deliver the same power. Standard AC voltmeters are calibrated to display this RMS value.

Q: Is this related to a 3-phase power supply?

Yes, absolutely! The given voltages represent a balanced 3-phase system. The individual voltages are the 'phase voltages,' and the differences between them are the 'line voltages.' In such a system, the line voltage is always $\sqrt{3}$ times the phase voltage.

Q: What would be the reading for $V_{ZX}$?

By symmetry, the potential difference $V_{ZX} = V_Z - V_X$ would also have the same magnitude, $\sqrt{3} V_0$, and its RMS value would be $V_0 \sqrt{\frac{3}{2}}$.

Question

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

Options:

(A) $(V_{YZ})_{	ext{rms}} = V_0 \sqrt{\frac{1}{2}}$

(B) $(V_{XY})_{	ext{rms}} = V_0 \sqrt{\frac{3}{2}}$

(C) independent of the choice of the two terminals

(D) $(V_{XY})_{	ext{rms}} = V_0$

Sigma Percentile · Accepted Answer

The problem of finding the potential difference between different terminals in an AC circuit can seem daunting if we rely solely on trigonometric identities. However, by translating the problem into the language of **phasors**, we can turn a messy algebraic calculation into an elegant geometric visualization.

### Visualizing the Phasors

In an AC circuit, voltages that vary sinusoidally with time can be represented as rotating vectors, or phasors. The length of the phasor represents the peak voltage, and its angle represents the phase.

Given the instantaneous voltages:
$$V_X = V_0 \sin \omega t$$
$$V_Y = V_0 \sin \left( \omega t + \frac{2\pi}{3} ight)$$
$$V_Z = V_0 \sin \left( \omega t + \frac{4\pi}{3} ight)$$

We can represent these as three phasors of length $V_0$, separated by $120^\circ$ ($\frac{2\pi}{3}$ radians) from each other. $V_X$ points at $0^\circ$, $V_Y$ points at $120^\circ$, and $V_Z$ points at $240^\circ$.

### The Potential Difference $V_{XY}$

We need to find the potential difference between terminals $X$ and $Y$, which is given by $V_{XY} = V_X - V_Y$.

Geometrically, subtracting a vector is the same as adding its negative. So, we take the phasor $V_Y$ and flip it by $180^\circ$ to get $-V_Y$. Since $V_Y$ is at $120^\circ$, $-V_Y$ will be at $120^\circ - 180^\circ = -60^\circ$.

Now, we add $V_X$ (at $0^\circ$) and $-V_Y$ (at $-60^\circ$). The angle between these two vectors is exactly $60^\circ$. Using the parallelogram law of vector addition, the magnitude of the resultant phasor $V_{XY}$ is:
$$|V_{XY}| = \sqrt{V_0^2 + V_0^2 + 2V_0^2 \cos 60^\circ}$$
$$|V_{XY}| = \sqrt{V_0^2 + V_0^2 + 2V_0^2 \left(\frac{1}{2}ight)} = \sqrt{3} V_0$$

### The Voltmeter Reading

An ideal AC voltmeter is designed to read the **RMS (Root Mean Square)** value of the voltage, not the peak value. The RMS value is related to the peak value by a factor of $\sqrt{2}$.

$$(V_{XY})_{	ext{rms}} = \frac{|V_{XY}|}{\sqrt{2}} = \frac{\sqrt{3} V_0}{\sqrt{2}} = V_0 \sqrt{\frac{3}{2}}$$

### The Potential Difference $V_{YZ}$

Let's repeat the process for terminals $Y$ and $Z$. The potential difference is $V_{YZ} = V_Y - V_Z$.

We flip $V_Z$ (which is at $240^\circ$) to get $-V_Z$ at $60^\circ$. The angle between $V_Y$ (at $120^\circ$) and $-V_Z$ (at $60^\circ$) is again $60^\circ$.

By the exact same geometric logic, the magnitude of $V_{YZ}$ is $\sqrt{3} V_0$, and its RMS value is:
$$(V_{YZ})_{	ext{rms}} = V_0 \sqrt{\frac{3}{2}}$$

### The Beauty of Three-Phase Systems

Notice that $(V_{XY})_{	ext{rms}}$ and $(V_{YZ})_{	ext{rms}}$ are identical. By symmetry, $(V_{ZX})_{	ext{rms}}$ would also be the same. This means the voltmeter reading is **independent of the choice of the two terminals**.

This setup is a perfect mathematical model of a **balanced 3-phase AC power system**, which is how electricity is generated and distributed worldwide. The individual voltages $V_X, V_Y, V_Z$ are the **phase voltages**, and the differences between them are the **line voltages**. As we've just proven, in a balanced star-connected system, the line voltage is always $\sqrt{3}$ times the phase voltage!

Select Answer:

Visualized Solution

Visualizing the Phasors

Potential Difference $V_{X Y}$

Calculating $V_{X Y}$

RMS Value of $V_{X Y}$

Potential Difference $V_{Y Z}$

Calculating $V_{Y Z}$

RMS Value of $V_{Y Z}$

Three-Phase System Analogy

The Sigma Insight: Alternating Current (AC) and Voltage

Visualizing the Phasors

The Potential Difference $V_{X Y}$

The Voltmeter Reading

The Potential Difference $V_{Y Z}$

The Beauty of Three-Phase Systems

Similar Questions

An alternating current is given by the equation $i = i_{1} sin ω t + i_{2} cos ω t$ . The rms current will be

An AC current is given by $I = I_{1} sin ω t + I_{2} cos ω t$ . A hot wire ammeter will give a reading

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

List-I

List-II

In an AC-circuit, an inductor, a capacitor and a resistor are connected in series with $X_{L} = R = X_{C}$ . Impedance of this circuit is

In an $L - C - R$ series AC circuit, the voltage across each of the components $L, C$ and $R$ is $50 V$ . The voltage across the $L - C$ combination will be

List-I

List-II

The alternating current is given by $i = {42 sin (\frac{2 π}{T} t) + 10} A$ The rms value of this current is ...... A.

An alternating voltage $V (t) = 220 sin 100 π t$ volt is applied to a purely resistive load of $50Ω$ . The time taken for the current to rise from half of the peak value to the peak value is

Select Answer:

Visualized Solution

Visualizing the Phasors

Potential Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VXY​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VXY​

RMS Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VXY​

Potential Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VYZ​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VYZ​

RMS Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VYZ​

Three-Phase System Analogy

The Sigma Insight: Alternating Current (AC) and Voltage

Visualizing the Phasors

The Potential Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VXY​

The Voltmeter Reading

The Potential Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VYZ​

The Beauty of Three-Phase Systems

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An alternating current is given by the equation i=i1​sinωt+i2​cosωt. The rms current will be

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An AC current is given by I=I1​sinωt+I2​cosωt. A hot wire ammeter will give a reading

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the given circuit, the AC source has ω=100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In an AC-circuit, an inductor, a capacitor and a resistor are connected in series with XL​=R=XC​. Impedance of this circuit is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In an L−C−R series AC circuit, the voltage across each of the components L,C and R is 50 V. The voltage across the L−C combination will be

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The alternating current is given by i={42​sin(T2π​t)+10}A The rms value of this current is ...... A.

Potential Difference $V_{X Y}$

Calculating $V_{X Y}$

RMS Value of $V_{X Y}$

Potential Difference $V_{Y Z}$

Calculating $V_{Y Z}$

RMS Value of $V_{Y Z}$

The Potential Difference $V_{X Y}$

The Potential Difference $V_{Y Z}$

An alternating current is given by the equation $i = i_{1} sin ω t + i_{2} cos ω t$ . The rms current will be

An AC current is given by $I = I_{1} sin ω t + I_{2} cos ω t$ . A hot wire ammeter will give a reading

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

In an AC-circuit, an inductor, a capacitor and a resistor are connected in series with $X_{L} = R = X_{C}$ . Impedance of this circuit is

In an $L - C - R$ series AC circuit, the voltage across each of the components $L, C$ and $R$ is $50 V$ . The voltage across the $L - C$ combination will be

The alternating current is given by $i = {42 sin (\frac{2 π}{T} t) + 10} A$ The rms value of this current is ...... A.