Question

An AC source operating at a frequency of 400 Hz has a maximum output voltage of 6.00 V. What is the smallest inductor that can be connected across the source and the rms current remain less than 2.50 mA? (Enter the inductance in H.)

Answer 1

L= 675.2 mH

Step-by-step explanation:

Given that

f= 400 Hz

Maximum voltage ,V(max)= 6 V

I(rms) = 2.5 mA

`I(max)=I(rms)√(2)`

`I(max)=(V(max))/(X_L)\\I(max)=(V(max))/(L* 2* \pi * f)`

`L=(V(max))/(I(max)* 2* \pi * f)`

Now by putting the values

`L=(6)/(2.5* 10^(-3)* \sqrt2 * 2* \pi * 400)`

L=0.6752

L= 675.2 m H

Inductance ,L= 675.2 mH