Solve RL Circuit Equation: Kirchhoff's Rule Explained

[Avichal]

Suppose a circuit with resistor R and inductor L with no source. I am trying to find kirchhoffs equation for this circuit - I am getting iR -Ldi/dt = 0 as my equation which is apparently wrong. I just cannot understand how do I make equations for such circuits.

 

[CWatters]

Should that be iR + Ldi/dt = 0

 

[Avichal]

Yes it should be but I don't get it why. Voltage across resistor is iR and then voltage across inductor decreases by Ldi/dt so iR-Ldi/dt=0

 

[vanhees71]

Draw your circuit and indicate an (arbitrary) direction, in which you want to count the current positive. Then use the right-hand rule to attach the surface-normal vector oriented positive relative to that direction of the current. Finally use Faraday's Law,
\partial_t \vec{B}=-\vec{\nabla} \times \vec{E},
and integrate (line integral) along the circuit in direction of the positve current. Then the left-hand side translates into L \frac{\mathrm{d} i}{\mathrm{d}t} for compact circuits, and the right-hand side you can transform into an integral along the surface, translating into -R i, where we have made use of Ohm's Law, \vec{E}=\vec{j}/\sigma. From this you get the desired equation,
L \frac{\mathrm{d} i}{\mathrm{d} t}=-R i.

 

[Avichal]

Sorry I am unaware of some of the things you said. Can you simplify please?