Chapter 21, Problem 22P

A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates in a 0.65-T magnetic field. The generator is used to power a lightbulb rated at 12.0V and 25.0 W. At what rate should the generator rotate to provide 12.0 V to the bulb? Consider the resistance of the wire on the armature.

To determine

The rate of the rotation of the generator to provide 12 V.

Answer to Problem 22P

Solution:

17 cycles per second.

Explanation of Solution

Given info:

$\begin{array}{l}\text{No}\text{.of Turns }\left(N\right)=85\\ \text{length }\left(l\right)\text{=6 cm}\\ \text{ =0}\text{.060 m}\\ \text{Diameter}=\text{0}\text{.59 mm}\\ \text{ =5}\text{.9}\times {\text{10}}^{-4}\text{ m}\\ \text{magnetic field}\left(B\right)\text{=0}\text{.65 T}\\ \text{Power}\left(P\right)\text{=25 W}\\ {\text{Voltage in bulb, V}}_{rms}\text{=12}\text{.0 V}\end{array}$

Formula to be used:

The internal resistance:

The resistance of uniform metallic conductor is given by

$R=\frac{\rho l}{A}$ [$l$ =length, $A$=area, $\rho$= electrical resistivity]

Calculation:

Applying values,

$\begin{array}{l}\text{R=}\frac{\left(1.6\times {10}^{-8}\right)\left(20.4\right)}{\pi {\left(\frac{\text{5}\text{.9}\times {\text{10}}^{-4}}{2}\right)}^2}\\ R=1.194\Omega \end{array}$ [Resistivity of copper is $\text{1}{\text{.6×10}}^{\text{-8}}\text{ Ωm}$]

Power of the light bulb rated at 12.0 V and 25 W.

Formula used:

By using power formula,

$P=I_{rms}{V}_{rms}$

Calculation:

$\begin{array}{l}\Rightarrow I_{rms}=\frac{P}{V_{rms}}\\ I_{rms}=\frac{25.0}{12.0}\\ I_{rms}=\text{2}\text{.08 A}\end{array}$

Voltage drop inside the generator,

$V=IR$

Applying values,

$\begin{array}{l}{V}_{drop}=\left(2.083\right)\left(1.194\right)\\ V_{drop}=\text{2}\text{.487 V}\end{array}$

Therefore, the actual Emf required = $\text{Voltage in bulb}$+ $V_{drop}$

$=12+2.487$

$\text{=14}\text{.49 }\left(\text{Approximately}\right)$

The above value is rms value of emf

The maximum EMF (peak voltage)is given by,

$\begin{array}{l}{E}_{peak}=\sqrt{2}{E}_{rms}\\ \text{ =}\sqrt{\text{2}}\text{×14}\text{.49}\\ \text{ =20}\text{.49 V}\end{array}$

For generator, by Faraday’s law,

Formula used:

$\begin{array}{l}{E}_{peak}=NBA\omega \\ E_{peak}=NBA\left(2\pi f\right)\\ \Rightarrow f=\frac{E_{peak}}{2\pi NBA}\end{array}$

Calculation:

Applying values,

$\begin{array}{l}f=\frac{20.49}{2\pi \left(85\right)\left(065\right)\left(0.06\right)}\\ f=\text{17 Hz }\left(\text{Approximately}\right)\end{array}$