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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.

Respuesta :

Answer:

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

Explanation:

We have given angular frequency [tex]\omega =850rad/sec[/tex]

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

Part A

We have given inductance [tex]L=10^{-2}H[/tex]

So inductive reactance [tex]X_L=\omega L=850\times 10^{-2}=8.5ohm[/tex]

So current [tex]i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A[/tex]

Part B

We have given inductance [tex]L=1 H[/tex]

So inductive reactance [tex]X_L=\omega L=850\times 1=850ohm[/tex]

So current [tex]i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA[/tex]

Part C

We have given inductance [tex]L=100 H[/tex]

So inductive reactance [tex]X_L=\omega L=850\times 100=85000ohm[/tex]

So current [tex]i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA[/tex]

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