- #1

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dw=Eidt where E is the emf of the variable voltage source.We put E=+L (di/dt).

can you please tell me why we put that value?? (i know the induced e.m.f on the inductor is - L (di/dt)

thanks

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- Thread starter zero_kilo
- Start date

- #1

- 10

- 0

dw=Eidt where E is the emf of the variable voltage source.We put E=+L (di/dt).

can you please tell me why we put that value?? (i know the induced e.m.f on the inductor is - L (di/dt)

thanks

- #2

rl.bhat

Homework Helper

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So dw = Eidt = L*i*di/dt*dt = L*i*di.

Now take integrations to get the energy stored in the inductor.

- #3

- 10

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So dw = Eidt = L*i*di/dt*dt = L*i*di.

Now take integrations to get the energy stored in the inductor.

thanks but this is what i actually asked

We put E=+L (di/dt).

can you please tell me why we put that value??

thanks

why we put E=+L dI/dt for the applied e.mf

- #4

- 548

- 2

why we put E=+L dI/dt for the applied e.mf

Consider the equation E = -L di/dt. The minus sign in the equation indicates that the induced EMF has an orientation that opposes the the change in current.

The reason why it's positive for your given equation, is because positive work must be done to induce the current. We need only the magnitude of E then.

- #5

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Consider the equation E = -L di/dt. The minus sign in the equation indicates that the induced EMF has an orientation that opposes the the change in current.

The reason why it's positive for your given equation, is because positive work must be done to induce the current. We need only the magnitude of E then.

thanks but why to drive the current through the circuit so as to defeat the induced e.m.f -L di/dt we need Ldi/dt; dont we not need a greater value??

thanks again.

- #6

ideasrule

Homework Helper

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- #7

- 548

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The induced emf is Ldi/dt; the negative sign just indicates it opposes the battery's emf.

I don't think that this is quite correct.

Consider a circuit with an applied emf, resistor, and inductor, where the current varies with time. If the current is increasing (di/dt > 0 ), then the self-induced emf opposes the direction of the applied emf. If the current is decreasing (di/dt < 0), the self-induced emf is actually oriented in the same direction as the applied emf.

In both cases, the self-induced emf has orientation that always opposes the "change in current", not the "applied emf."

- #8

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thanks :-)

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