Q77P In Fig. 31-38, a three-phase gen... [FREE SOLUTION]

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Chapter 31: Q77P (page 940)

In Fig. 31-38, a three-phase generator G produces electrical power that is transmitted by means of three wires. The electric potentials (each relative to a common reference level) are $V_{1} = A sin 蝇_{d} t$ for wire 1, $V_{2} = A sin (蝇_{d} t - 120^{0})$ for wire 2, and $V_{3} = A sin (蝇_{d} t - 240^{0})$ for wire 3. Some types of industrial equipment (for example, motors) have three terminals and are designed to be connected directly to these three wires. To use a more conventional two-terminal device (for example, a lightbulb), one connects it to any two of the three wires. Show that the potential difference between any two of the wires (a) oscillates sinusoidally with angular frequency $蝇_{d}$ and (b) has an amplitude of $A \sqrt{3}$ .

Short Answer

Expert verified

a. The potential difference between any two of the wires oscillates sinusoidally with the angular velocity $蝇_{d}$ .

b. The potential difference between any two of the wires has an amplitude of $A \sqrt{3}$ .

Step by step solution

Step 1: Given Information

. $V_{1} = A sin 蝇_{d} t$
. $V_{2} = A sin (蝇_{d} t - 120^{0})$
. $V_{3} = A sin (蝇_{d} t - 240^{0})$

Determining the concept

The three-phase generator has three ac voltages that are $120 掳$ out of phase with each other. Calculate the potential difference between any two wires by using the trigonometric identity.

The formula is as follows:

$A - \sin B = 2 \sin [\frac{A - B}{2}] \cos [\frac{A + B}{2}]$

(a) Determining the potential difference between any two of the wires oscillating sinusoidally with the angular frequency ωd

Consider different combinations for potential difference:

$n V_{12} = V_{1} - V_{2} n V_{12} = A sin 蝇_{d} t - A sin (蝇_{d} t - 120^{0})$

By using the trigonometric identity,

$n V_{12} = 2 A \sin (\frac{蝇_{d} t - 蝇_{d} t + 120 掳}{2}) \cos (\frac{蝇_{d} t + 蝇_{d} t - 120 掳}{2}) n V_{12} = 2 A \sin (60 掳) \cos (蝇_{d} t - 60 掳) n V_{12} = 2 A (\sqrt{\frac{3}{2}}) 蝇_{d} t - 60 掳 n V_{12} = \sqrt{3} A (蝇_{d} t - 60 掳)$

Now,

$n V_{13} = V_{1} - V_{3} n V_{13} = A sin 蝇_{d} t - A sin (蝇_{d} t - 240^{0})$ .

By using the trigonometric identity,

role="math" localid="1663157004430" $n V_{13} = 2 A \sin (\frac{蝇_{d} t - 蝇_{d} t + 240 掳}{2}) \cos (\frac{蝇_{d} t + 蝇_{d} t - 240 掳}{2}) n V_{13} = 2 A \sin (120 掳) \cos (蝇_{d} t - 120 掳) n V_{13} = 2 A (\sqrt{\frac{3}{2}}) \cos (蝇_{d} t - 120 掳) n V_{13} = \sqrt{3} A \cos (蝇_{d} t - 120 掳)$

Similarly,

$n V_{23} = V_{2} - V_{3} n V_{23} = A sin (蝇_{d} t - 120^{0}) - A sin (蝇_{d} t - 240^{0})$

By using the trigonometric identity,

$n V_{23} = 2 A \sin (\frac{蝇_{d} t - 120 掳 - 蝇_{d} t + 240 掳}{2}) \cos (\frac{蝇_{d} t - 120 掳 + 蝇_{d} t - 240 掳}{2}) n V_{23} = 2 A \sin (60 掳) \cos (蝇_{d} t - 180 掳) n V_{23} = 2 A (\sqrt{\frac{3}{2}}) \cos (蝇_{d} t - 180 掳) n V_{23} = \sqrt{3} A \cos (蝇_{d} t - 180 掳)$

The expressions $n V_{12}, n V_{13}, n V_{23}$ are the sinusoidal function of $t$ angular frequency $蝇_{d}$ .

Hence, the potential difference between any two of the wires oscillates sinusoidally with the angular velocity $蝇_{d}$ .

(b) Determining the potential difference between any two of the wires has amplitude of A3

$n V_{12} = \sqrt{3} A (蝇_{d} t - 60^{0}) n V_{13} = \sqrt{3} A cos (蝇_{d} t - 120^{0}) n V_{23} = \sqrt{3} A cos (蝇_{d} t - 180^{0})$

Thus, all these expressions have an amplitude of $\sqrt{3} A .$

Hence, the potential difference between any two of the wires has an amplitude of $\sqrt{3} A .$

Showed that the potential difference between any two of the wires oscillates sinusoidally with the angular velocity $蝇_{d},$ and it has an amplitude of $\sqrt{3} A .$