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.
Where I've made (a) mistake(s)?
