(a) Circuit with ac source and inductor a

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Equation (attached) says that vab = L di/dt (see Fig). Using Faraday’s law, explain why point a is at higher potential than point b when i is in the direction shown and is increasing in magnitude. When i is counterclockwise and decreasing in magnitude, is vab = L di/dt still correct, or should it be vab = -L di/dt? Is vab = L di/dt still correct when i is clockwise and increasing or decreasing in magnitude? Explain

(a) Circuit with ac source and inductor
a
Transcribed Image Text:(a) Circuit with ac source and inductor a
Expert Solution
Step 1

Since the current flows from the higher potential to lower potential. And it is given that the current is flowing from, point a to point b. Therefore, point a will be at the higher potential.

Step 2

Now when the current will be in the counter-clockwise direction and decreasing di/dt will be less than zero or negative. Therefore, then the potential be given as,

Vab=-Ldidt

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