Question

A simple generator has a square armature 7.0 cm on a side. The armature has 75 turns of 0.59-mm diameter copper wire and rotates in a 0.800-T magnetic field. The generator is used to power a lightbulb rated at 12.0 V 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. (Express your answer to two significant figures and include the appropriate units.)

Answer 1

75 x 7X 4 :: square armature) emf generated Lao رہا W► from Faraday's we have the E = NABW where N number of twins A area of the armature angular frequency B magnetic field strength. let us determined the emf (E) to be converted to E = va x (drop in voltage of generator + voltage required for the bulb) The factor of re is taken as the voltage given are, peak voltages. Resistance of the wire R = where e resistivity ett wire el →( Ly length of the A,+ erossectional area of the wire A diameter •-radius R TAN 2 L = No. of twin x side length of the armature x 4 = L = ra 2 from (6) * L = 2100 cm = 21m 0.59 89 0.295 mm = 2.95 X10 ezlizx10-8am (for copper wire) Rs 1.7x10-8 x 21 1.3062 TT X (2.95% 10-499 resistance of the Voltage due to this armature agenerator (VI)= IR where I current supplied by the Power used by the bulb generator 25 of the bulb 22.083 A drop i, I= voltage al N275 B=0.8T 20.81 V = IR 32.083 X 1.306 2.72 Volt :.Ea 82x(Vit voltage for the bulb) = 52x (2.72 +12) 7 20.81 voit; A =(0.07) 6) E NAR 75X.10.07)% 0.8 70.78 nad/sec from ean = 71 rad/sec generator should rotate at arate of zinadfree Ans