Chapter 21, Problem 20P

To determine

(a)

The rms voltage output of the generator.

Answer to Problem 20P

Solution:

rms voltage of generator is 810 V.

Explanation of Solution

Given info:

Turns(N)$=550$

Magnetic field $\left(B\right)=\text{0}\text{.55 T}$

Diameter $\text{=8 cm}$

Radius $\text{=}\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$=\text{4 cm}$

$\text{=0}\text{.04 m}$

Frequency $f=\text{120 }\text{rev}/\text{s}$

Rms voltage, $V_{rms}=\text{150 V}$

Formula to be used:

The rms output, $V_{rms}=\frac{NB\omega A}{\sqrt{2}}$

[Derived from peak induced emf $E_{peak}=NB\omega A$]

Calculation:

Applying values

$V_{rms}=\frac{\left(550\right)\left(0.55\right)\left(2\pi f\right)\pi {\left(0.04\right)}^2}{\sqrt{2}}$$[\omega =2\pi f,A=\pi r^2]$

$V_{rms}=\text{809}\text{.84 V}$

$V_{rms}=\text{810 V }\left(\text{Approximately}\right)$

To determine

(b)

The rotation frequency.

Answer to Problem 20P

Solution:

Rotational frequency=240 rev/s.

Explanation of Solution

Given info:

$V_{rms}=\text{810 V }$

According to the rms output $V_{rms}=\frac{NB\omega A}{\sqrt{2}}$,the output voltage is directly proportional torotation of the frequency.

$V_{rms}\propto f$

So, to double the output voltage, the rotation frequency must double to 240 rev/s