Chapter 33, Problem 12SP

(a)

To determine

The effect on the induced emf of a generator if the flux per pole is doubled.

Answer to Problem 12SP

Solution:

Doubled

Explanation of Solution

Given data:

The flux per pole is doubled.

Formula used:

The induced emf of the generator is expressed as

$E=\left[NAB\left(2\pi f\right)\right]\cos \left(2\pi ft\right)$

Here, $B$ is the magnetic field, $A$ is the area of the winding, $N$ is the number of turns in the winding, $f$ is the frequency, and $t$ is the time.

Explanation:

Consider the expression for induced emf of the generator at its initial condition:

$E_i=\left[NA{B}_i\left(2\pi f\right)\right]\cos \left(2\pi ft\right)$ …… (1)

Consider the expression for induced emf of the generator when flux per pole is doubled:

$E_f=\left[NA{B}_f\left(2\pi f\right)\right]\cos \left(2\pi ft\right)$ …… (2)

From equation (1) and (2),

$\begin{array}{c}\frac{E_i}{E_f}=\frac{\left[NA{B}_i\left(2\pi f\right)\right]\cos \left(2\pi ft\right)}{\left[NA{B}_f\left(2\pi f\right)\right]\cos \left(2\pi ft\right)}\\ =\frac{B_i}{B_f}\end{array}$

Understand that the magnetic flux per pole is doubled. Hence, magnetic field should be doubled.

Substitute $2B_i$ for $B_f$

$\begin{array}{c}\frac{E_i}{E_f}=\frac{B_i}{2B_i}\\ =\frac{1}{2}\end{array}$

Solve for $E_f$

$E_f=2E_i$

Conclusion:

Therefore, if the flux per pole is doubled, the induced emf of a generator doubles as well.

(b)

To determine

The effect on the induced emf of a generator if the speed of the armature is doubled.

Answer to Problem 12SP

Solution:

Doubled

Explanation of Solution

Given data:

The speed of the armature is doubled.

Formula used:

The induced emf of the generator is expressed as

$E=\left[NAB\left(2\pi f\right)\right]\cos \left(2\pi ft\right)$

Here, $B$ is the magnetic field, $A$ is the area of the winding, $N$ is the number of turns in the winding, $f$ is the frequency, and $t$ is the time.

Explanation:

Consider the expression for induced emf of the generator at its initial condition:

$E_i=\left[NAB\left(2\pi f_i\right)\right]\cos \left(2\pi ft\right)$ …… (3)

Consider the expression for induced emf of the generator when the speed of the armature is doubled, assuming the doubling of speed will not have much effect on the cosine term of the induced emf for small speed. Therefore,

$E_f=\left[NAB\left(2\pi f_f\right)\right]\cos \left(2\pi ft\right)$ …… (4)

From equation (3) and (4),

$\begin{array}{c}\frac{E_i}{E_f}=\frac{\left[N_{i}AB\left(2\pi f_i\right)\right]\cos \left(2\pi ft\right)}{\left[N_{f}AB\left(2\pi f_f\right)\right]\cos \left(2\pi ft\right)}\\ =\frac{f_i}{f_f}\end{array}$

Substitute $2f_i$ for $f_f$

$\begin{array}{c}\frac{E_i}{E_f}=\frac{f_i}{2f_i}\\ =\frac{1}{2}\end{array}$

Solve for $E_f$

$E_f=2E_i$

Conclusion:

Therefore, if the speed of the armature is doubled, the induced emf of a generator will also be double.