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

Answer:

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

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 $X_L=\omega L=850\times 10^{-2}=8.5ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A$

Part B

We have given inductance $L=1 H$

So inductive reactance $X_L=\omega L=850\times 1=850ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA$

Part C

We have given inductance $L=100 H$

So inductive reactance $X_L=\omega L=850\times 100=85000ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA$