Question

An a.c. generator consists of a coil of 50 turns and area 2.5 m² rotating at an angular speed of 60
rad/s in uniform magnetic field B = 0.30 T between two fixed pole pieces. The resistance of the
circuit including the coil is 500 2. What is the maximum current drawn by the generator?​

Answer 1

Answer:

Explanation:

hiii answer is given in the picture

Answer 2

The maximum current drawn by the generator is 4.5 A.

Explanation:

It is given that,

Number of turns, N = 50

Area of the coil, $A=2.5\ m^2$

Angular velocity, $\omega=60\ rad/s$

Magnetic field, B = 0.3 T

Resistance, R = 500 ohms

Due to the rotation of the coil, an emf is induced is given by :

$\epsilon=NBA\omega$

$\epsilon=50\times 0.3\times 2.5\times 60$

$\epsilon=2250\ V$

Let I is the maximum current drawn by the generator. It can be calculated using Ohm's law as :

$\epsilon=IR$

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

$I=\dfrac{2250}{500}$

I = 4.5 A

So, the maximum current drawn by the generator is 4.5 A. Hence, this is the required solution.

Learn more :

Induced emf

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