1. Aug 5, 2016

### Clara Chung

1. The problem statement, all variables and given/known data
On the right bottom side ,when the switch is closed the equation is given V(EG)= +V+V(FG)
The photo is attached

2. Relevant equations
Concept

3. The attempt at a solution
What is the difference between the back emf in the inductor and the p.d across the inductor? I think the current flow from E to G, the p.d. due to its resistance should be in opposite sign with the back emf in the inductor, but they are in the same sign as shown. Please explain.

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2. Aug 5, 2016

### cnh1995

When switch is closed, current flows from E to G. The non-ideal inductor can be modelled as an ideal inductor in series with its resistance.
The difference is in the directions. Voltage drop is numerically equal to the back emf of the inductor. Their directions are opposite. If voltage across inductor is 5V at a particular instant, that means 5V from the source are dropped across the inductor and its back emf at that instant is 5V.

3. Aug 5, 2016

### andrevdh

I tend to think that those two equitions need to be swopped around or alternatively they work if one is to move around the circuit in the anticlockwise direction, that is VGE, which is how they voltages would add up positively if one move up against the current.

4. Aug 5, 2016

### TSny

Yes.

Perhaps VGE represents the potential at G relative to E. So, VGE is the change in potential if you start at E and move to G.

VEG is the change in potential if you start at G and move to E. The equations appear to be written in terms of VEG. The equations appear to me to be correct as written in the text.

5. Aug 6, 2016

### Staff: Mentor

As cnh1995 pointed out, a practical inductor can be modelled as an ideal inductance in series with a resistance. So when you measure voltage across the terminals of a practical inductor you are measuring the sum of two voltages: voltage due to inductance, $\color{blue}{L\dfrac{di}{dt}}$ + Ohmic drop, $\color{blue}{i{\cdot}R}$.