Question

The armature of a 60 Hz ac generator rotates in a 0.12 T B-field. If the area of the coils is 0.03 m2, how many loops must the coil contain if the peak output is to be V = 270 V? What will the current & power output (not the power loss) be if the resistance of the coil is 0.4 ??

Answer 1

Explanation:

it is given that,

Frequency of AC generator, f = 60 Hz

Magnetic field, B = 0.12 T

Area of the coil, A = 0.03 m²

Peak out put, V = 270 V

Resistance of the coil, R = 04 ohms

1. Let N is the number of loops the coil must contain. The peak voltage of AC generator is given by :

$V=NBA\omega$

$N=\dfrac{V}{BA\omega}$

$N=\dfrac{270}{0.12\times 0.03\times 2\pi \times 60}$

N = 199

2. Applying Ohm's law as :

$I=\dfrac{V}{R}$

$I=\dfrac{270}{0.4}$

I = 675 A

3. Power, $P=V\times I$

$P=270\times 675$

P = 182250 watts

Hence, this is the required solution.