Please further explain this formula about voltage

[Clara Chung]

Homework Statement

On the right bottom side ,when the switch is closed the equation is given V(EG)=

+V+V(FG)
The photo is attached

Homework Equations

Concept

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.

 

[cnh1995]

When switch is closed, current flows from E to G. The non-ideal inductor can be modeled as an ideal inductor in series with its resistance.

What is the difference between the back emf in the inductor and the p.d across the inductor?

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.

 

[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 V_GE, which is how they voltages would add up positively if one move up against the current.

 

[TSny]

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 V_GE, which is how they voltages would add up positively if one move up against the current.

Yes.

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

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

 

[NascentOxygen]

What is the difference between the back emf in the inductor and the p.d across the inductor?

As cnh1995 pointed out, a practical inductor can be modeled 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, $\color{blue}{i{\cdot}R}$.