Question

An inductor with L = 9.80 mH is connected across an ac source that has voltage amplitude 41.5 V . What value for the frequency of the source results in a current amplitude of 4.50 A ?

Answer 1

Given data:

Inductance of coil (L)= 9.80 mH = 9.8 x 10^-3 H

Voltage amplitude of the ac source (V)= 41.4 V

Now we have to find out the frequency of source which will result in current amplitude of 4.50 A

We know that inductive reactance (resistance offered by the inductor) X_L is given by:

Where L is the inductance of inductor

is the frequency of the source

Now using the Ohm's law :

Where V is the applied voltage

I is the current

and R is the resistance (X_L in this case)

Here we have inductive reactance which is equivalent of resistance in the circuit.

Substituting the values in the Ohm's law we get:

Substituting this value of the reactance in previous equation for X_L:

Which is the required value of the frequency of source needed to get current amplitude of 4.5 A

Please give a up vote if you like the answer.

If you dislike the answer then put your queries, feedback, or suggestion in comments section.

If it doesn't gets resolved then down vote the answer.