Question

The resistanceless inductor is connected across the ac source whose voltage amplitude is 24.5 V and angular frequency is 850 rad/s . Find the current amplitude if the self-inductance of the inductor is: Part A: 1.00×10−2 H . Answer in A.

Part B: 1.00 H . Answer in mA.
Part C: 100 H . Answer in mA.

Answer 1

Part A 2.88 A, PART B 28.82 mA PART C 0.288 mA

Step-by-step explanation:

We have given angular frequency
`\omega =850rad/sec`

Voltage source has an amplitude of 24.5 V, So V= 24.5 volt

Part A

We have given inductance
`L=10^(-2)H`

So inductive reactance
`X_L=\omega L=850* 10^(-2)=8.5ohm`

So current
`i=(V)/(X_L)=(24.5)/(8.5)=2.8823A`

Part B

We have given inductance
`L=1 H`

So inductive reactance
`X_L=\omega L=850* 1=850ohm`

So current
`i=(V)/(X_L)=(24.5)/(850)=28.82mA`

Part C

We have given inductance
`L=100 H`

So inductive reactance
`X_L=\omega L=850* 100=85000ohm`

So current
`i=(V)/(X_L)=(24.5)/(85000)=0.288mA`