Chapter 32, Problem 44PQ

(a)

To determine

The graph of $\epsilon$ versus $t$ for each coil.

Answer to Problem 44PQ

Plot for the induced emf in coil $1$ is shown below.

Plot for the induced emf in coil 2 is shown below.

Plot for induced emf in coil $3$ is shown below.

Explanation of Solution

As it is known that a three phase generator has $3$ coils wound over a common armature, in away that each coil is set at an angle of $120°$ to the other two.

Therefore, a phase a difference of $120°$ occurs between $1$, $\text{2}$ and $2$, $3$ and a phase difference of $240°$ between $1$, $3$

Write the expression to calculate the induced emf in the coil.

    $\epsilon ={\epsilon }_0\sin \left(\omega t+\varphi \right)$                               (I)

Here, $\epsilon$ is the induced emf, ${\epsilon }_0$ is the peak value of the induced emf, $\omega$ is the angular velocity and $\varphi$ is  the initial phase of the coil when time is zero.

Write the expression to calculate the peak value of emf.

    $\epsilon =\left(\sqrt{2}\right){\epsilon }_{\max }$                                  (II)

Here, ${\epsilon }_{\max }$ is the rms value of the emf

Write the expression for the angular frequency.

    $\omega =2\pi f$                        (III).

Here, $f$ is the frequency.

Let the coil $1$ be set such that when time is zero the emf induced is also zero.

Substitute $0$ for $\epsilon$ and $0$ for $t$ in Equation (I).

    $\begin{array}{c}0={\epsilon }_0\sin \left(0+\varphi \right)\\ \sin \varphi =0\\ \varphi =0\end{array}$

Write the Equation for the emf of coil $1$.

    $\epsilon ={\epsilon }_0\sin \omega t$                               (IV).

Convert degree in to radian

    $\begin{array}{c}120°=120°\times \frac{\pi }{180°}\text{rad}\\ \text{=}\frac{\text{2}\pi }{3}\text{rad}\end{array}$

Write the expression for induced emf in coil $2$.

Substitute $\frac{\text{2}\pi }{3}\text{rad}$ for $\varphi$ in Equation (I).

    $\epsilon ={\epsilon }_0\sin \left(\omega t+\frac{\text{2}\pi }{3}\text{rad}\right)$                                 (V)

Convert degree in to radian

    $\begin{array}{c}240°=240°\times \frac{\pi }{180°}\text{rad}\\ \text{=}\frac{\text{4}\pi }{3}\text{rad}\end{array}$

Write the expression for induced emf in coil $3$.

Substitute $\frac{\text{4}\pi }{3}\text{rad}$ for $\varphi$ in Equation (I).

    $\epsilon ={\epsilon }_0\sin \left(\omega t+\frac{\text{4}\pi }{3}\text{rad}\right)$                                 (VI)

Calculate the peak emf

Substitute $110\text{V}$ for ${\epsilon }_{\max }$ in Equation (II)

    $\begin{array}{c}\epsilon =\left(\sqrt{2}\right)110\text{V}\\ =156\text{V}\end{array}$

Conclusion:

Calculate the angular frequency.

Substitute $60\text{Hz}$ for $f$ in Equation (III).

    $\begin{array}{c}\omega =2\pi \times 60\text{Hz}\\ \text{=377}\text{rad}/\text{sec}\end{array}$

Calculate the equation of emf for coil $1$.

Substitute $156\text{V}$ for ${\epsilon }_0$ and $377\text{rad}/\text{sec}$ for $\omega$ in Equation (IV).

    $\epsilon =156\text{V}\sin \left(377\text{rad}/\text{sec}\times t\right)$                             (VII)

Calculate the equation of emf for coil $2$.

Substitute $156\text{V}$ for ${\epsilon }_0$ and $377\text{rad}/\text{sec}$ for $\omega$ in Equation (V).

    $\epsilon =156\text{V}\sin \left(377\text{rad}/\text{sec}\times t+\frac{\text{2π}}{3}\text{rad}\right)$                 (VIII)

Calculate  the equation of emf for coil $3$.

Substitute $156\text{V}$ for ${\epsilon }_0$ and $377\text{rad}/\text{sec}$ for $\omega$ in Equation (IV).

    $\epsilon =156\text{V}\sin \left(377\text{rad}/\text{sec}\times t+\frac{\text{4π}}{3}\text{rad}\right)$                  (VI)

Draw the table to plot the graph at different time interval.

Plot for the induced emf in coil $1$ is shown below.

Plot for the induced emf in coil 2 is shown below.

Plot for induced emf in coil $3$ is shown below.

(b)

To determine

The ways in which a three phase generator is better than a one phase generator.

Answer to Problem 44PQ

The three phase generator has higher efficiency and higher power factor compared to a one phase generator which is its major advantage over one phase generator.

Explanation of Solution

The advantage of three phase generator over a one phase generator are as follows.

1. 1. The power output of a three phase generator is constant whereas the power generated by a one phase generator is pulsating that it is not constant and behaves in an alternating manner.
2. 2. The three phase generator has higher power factor as compared to one phase generator.
3. 3. The three phase generator has much higher efficiency as compared to the one phase motor.