Say I have the inductor in the following circuit with certain current already in the circuit flowing in the specified direction:


simulate this circuit – Schematic created using CircuitLab

If I had to write KVL for this circuit, would I write $$ L{di}/{dt}-iR=0$$ $$or$$ $$ L{di}/{dt}+iR=0$$

I always get super confused about what sign to take for the induction. I know by intuition(by the fact that the solution for the differential equation of (1) would be an exponent with positive power which is absurd) that it should be (2), but I always get stuck on that sign.


I'd use KCL (not KVL) with the bottom node "grounded" and the top node as \$V\$:

$$\frac{V}{R}+\frac1{L}\int V\:\text{d}t=0\:\text{A}$$

Taking the derivative with respect to time:


The integrating factor is then \$\mu=e^{^{\int\frac{R}{L}\text{d}t}}=e^{^{\frac{R}{L}t}}\$ and the solution is:

$$\begin{align*}V&=e^{^{-\frac{R}{L}t}}\int0\:\text{A}\cdot e^{^{\frac{R}{L}t}}\:\text{d}t\\\\&=e^{^{-\frac{R}{L}t}}\left(0\:\text{V}+C_0\right)\\\\&=C_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$

From initial conditions, we know that \$C_0=I_0\cdot R\$. So:

$$\begin{align*}V_t&=I_0\cdot R\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$

Here, \$I_0\$ is the initial current in the specified direction.

Using your approach, let's start with a schematic:


simulate this circuit – Schematic created using CircuitLab

The above schematic will help you keep the signs straight, I think. Note that \$V=L\,\frac{\text{d} I}{\text{d} t}\$ where \$V\$ here is the difference between the top of \$L\$ and the bottom of \$L\$ when the current in \$L\$ is positive when it is down in direction. So in the above case, we find that \$\left(0\:\text{V}-V\right)=L\,\frac{\text{d} I}{\text{d} t}\$ or \$V=-L\,\frac{\text{d} I}{\text{d} t}\$. So,

$$\begin{align*} 0\:\text{V}-L\,\frac{\text{d} I}{\text{d} t} - I\cdot R&=0\:\text{V}\\\\ \frac{\text{d} I}{\text{d} t} + I\cdot \frac{R}{L}&=0\:\text{V} \end{align*}$$

The integrating factor (again) is then \$\mu=e^{^{\int\frac{R}{L}\text{d}t}}=e^{^{\frac{R}{L}t}}\$ and the solution is:

$$\begin{align*}I&=e^{^{-\frac{R}{L}t}}\int0\:\text{V}\cdot e^{^{\frac{R}{L}t}}\:\text{d}t\\\\&=e^{^{-\frac{R}{L}t}}\left(0\:\text{A}+C_0\right)\\\\&=C_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$

From initial conditions, we know that \$C_0=I_0\$. So:

$$\begin{align*}I_t&=I_0\cdot e^{^{-\frac{R}{L}t}}\end{align*}$$

Here again, \$I_0\$ is the initial current in the specified direction, with positive values being down.

I don't like this approach quite so much, in this case, because you have to "think more" about the signs. It works fine. It's just slightly more "twisting of the mind" to make sure everything is fine. The KCL approach is much easier to keep straight because all you do is make sure that you put the outgoing currents on the left and the incoming currents on the right (or visa versa.) And at least to me, that's easier to keep track of.

  • \$\begingroup\$ This does make things clearer, thanks! But if I were to be adamant about using KVL, what would I take across the inductor? When is the voltage across inductor taken as Ldi/dt and when is it -Ldi/dt? \$\endgroup\$ – user_9 Jan 14 at 15:45
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    \$\begingroup\$ @user_9 It's the same problem and the results and approach are quite similar. In fact, you should be able to use your approach using the above process to find the solution for current, instead of voltage. I wanted to choose a different path in order to shake up your thinking a little bit. But not too say your approach doesn't work. It does. \$\endgroup\$ – jonk Jan 14 at 16:12
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    \$\begingroup\$ @user_9 If you want, I can easily stick with the way you started out and just solve that, instead. You have the right starting approach. Just follow through with it. Your sign is actually correct. Do you want me to spend time talking about how to ensure your signs are correct? Because if so, that's also why I chose KCL. It is easier to get the signs right, that way. \$\endgroup\$ – jonk Jan 14 at 16:14
  • \$\begingroup\$ This really helped clear things up, thanks! \$\endgroup\$ – user_9 Jan 14 at 16:57

When learning to analyze circuits it is always a good idea to explicitly mark the assumed orientation of voltages and assumed direction of currents on the schematic. For your example, the passive sign convention says that current enters the element at the terminal that has the more positive voltage.

So, the positive ends of the voltages across the two elements would be at the bottom end of the inductor and the top end of the resistor. Now when you write KVL it is clear that the equation should be

$$V_L + V_R = L\frac{di}{dt} + Ri = 0$$

If you want to put the positive voltage terminal at the top of both elements, then the assumed direction of current and the assumed polarity of voltage do not follow the passive sign convention for the inductor, and the voltage across the inductor would now be \$V'_L = -L\frac{di}{dt}\$. So the KVL equation becomes

$$-V'_L + V_R = -(-L\frac{di}{dt}) + Ri = L\frac{di}{dt} + Ri = 0$$

The result makes sense in this case because the current is decreasing from its initial value. So, \$\frac{di}{dt} < 0\$ and the voltage across the inductor, when observing the passive sign convention, is negative. If you assume instead that the inductor voltage is positive on the inductor's top terminal then this new inductor voltage would be positive, and it makes sense that the inductor is in parallel with a resistor that has a positive voltage.


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