# Direction of magnetic flux of mutual inductance where coupling coefficient is almost 1

I've drawn the below diagram of the circuit.

We will handle the system as the switch is closed .

The coupling coefficient of the system is almost 1

I assumed $$\~ L_{1} ,L_{2},M ~\$$ are all positive real numbers .

The book showed me the below equations.

$$L_{1} \frac{ dI_{1} }{ dt } +M \frac{ dI_{2} }{ dt }+R_{1}I_{1} =E \tag{1}$$

$$L_{2} \frac{ dI_{2} }{ dt } +M \frac{ dI_{1} }{ dt }+R_{2}I_{2} =0 \tag{2}$$

What I thought for understand the above 2 equations are as below .

Firstly I focused on the left side circuit .

$$\Phi_{1} :=\text{magnetic flux at coil1 made by } ~I_{1}~\text{only}$$

$$\Phi_{2} :=\text{magnetic flux made at coil2 by } ~ \Phi_{1} ~$$

As I apply KVL .

$$\underbrace{L_{1} \frac{ dI_{1} }{ dt }}_\text{voltage drop} + \underbrace{M \frac{ dI_{2} }{ dt }}_\text{why voltage drop?} + \underbrace{R_{1}I_{1}}_\text{voltage drop} - \underbrace{E}_\text{voltage rise} = 0 ~~ \leftarrow~~ \text{Transformed eqn1}$$

$$\~ \Phi_{1} ~\$$ makes a direction of loop of clockwise.

Since $$\~ \Phi_{1} ~\$$ enters coil2 from top to bottom , $$\~ \Phi_{2} ~\$$ should flows from bottom to top . Then the top side endpoint of coil2 takes higher potential and the bottom endpoint takes a lower potential . Hence the symbol of direction of $$\~ I_{2} ~\$$ in the diagram is adequate .

$$\~ \Phi_{2} ~\$$ makes a direction of coutnerclockwise.

$$\Phi_{1,2} :=\text{magnetic flux made at coil1 by } ~ \Phi_{2}$$

The direction of $$\~ \Phi_{1,2} ~\$$ is clockwise.

The top side endpoint of the coil1 takes a higher potential and the lower endpoint takes a lower potential.

So I think the negative sign should be attached to $$\~ M \frac{ dI_{2} }{ dt } ~\$$ to make an equation of KVL of left side circuit.