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Phasor diagram - why phasor A leads phasor B?

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phasor diagram of inductor

Say A is proportional to the time derivative of B, e.g. Faraday's law states that induced emf E = dΦ/dt = Ldi/dt. When A and B are represented in phasor diagram, why phasor A leads phasor B? Anything to do with 'mathematic' theory?

Thanks
 

phasor diagram jwl

In an ac circuit, current varies as a sinusoidal wave. For this variation of current, rate of change of current and hence induced emf (=-L(di/dt)) will be maximum at the time of start, and hence the induced emf will be maximum. However, due to negaive sign in the expression for induced emf (which takes care of Lenz's law), it will be lag behind the current in phase by Pi/2.
 

nonlinear said:
In an ac circuit, current varies as a sinusoidal wave. For this variation of current, rate of change of current and hence induced emf (=-L(di/dt)) will be maximum at the time of start, and hence the induced emf will be maximum. However, due to negaive sign in the expression for induced emf (which takes care of Lenz's law), it will be lag behind the current in phase by Pi/2.
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I thought the current flowing through an inductor will lag behind the voltage across the inductor. Pls correct me if I'm wrong.
 

I thought the current flowing through an inductor will lag behind the voltage across the inductor. Pls correct me if I'm wrong.
Click to expand...

You are correct. We can see this from a purely circuits perspective as well. Using Ohm's Law, we know V = IZ, where Z is the impedence. For an inductor Z = jwL. So V = I*(jwL).

Let's find the phase. arg(V) = arg(I*(jwL)). The phase of a product is equal to the sum of the phases, so arg(V) = arg(I) + arg(jwL).

arg(V) = arg(I) + pi/2. It is now apparent that the voltage across an inductor leads the current by pi/2.
 

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    powersys

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jayc said:
I thought the current flowing through an inductor will lag behind the voltage across the inductor. Pls correct me if I'm wrong.
Click to expand...

You are correct. We can see this from a purely circuits perspective as well. Using Ohm's Law, we know V = IZ, where Z is the impedence. For an inductor Z = jwL. So V = I*(jwL).

Let's find the phase. arg(V) = arg(I*(jwL)). The phase of a product is equal to the sum of the phases, so arg(V) = arg(I) + arg(jwL).

arg(V) = arg(I) + pi/2. It is now apparent that the voltage across an inductor leads the current by pi/2.
Click to expand...
Why arg(jwL) is equal to pi/2? Thanks.
 

Why arg(jwL) is equal to pi/2? Thanks.
Click to expand...

jwL is a purely imaginary value (since w and L are both real), so by definition, the phase is pi/2.
 

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    powersys

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