Recently I had a reader comment with a request for an example of Kirchoff’s Voltage Law (KVL) with Ohm’s Law. KVL and Ohm’s Law are both used in the circuit analysis methodology called Mesh Circuit Analysis. Consequently I thought I would analyze a small circuit and demonstrate the general principles of this technique that can be used on any circuit.
The circuit we will use below has a 10 volt source and 4 resistors configured as seen.
The first step is to label the circuit as I have done in red ink. This includes the current flowing clockwise through each loop (or mesh) and the voltage drops across each passive element. (In this case all the passive elements are resistors.) Next we will use the picture to write some equations.
If you look at the diagram at the first loop and start at the top, then travel around the loop going clockwise and add up the voltage drops you get the first equation below. Notice the voltage drop across the source is negative because it is not a drop it is an increase.Also notice the sum must equal zero.
The next 2 lines are simply writing the unknown voltage drops in terms of Ohm’s Law. The voltage drop across the resistor is equal to the current through the resistor times the resistance. Notice for resistor 2 the currents are opposing one another. In other words current 1 runs from the top of the circuit toward the bottom of the circuit through resistor 2. Current 2 runs from the bottom to the top. That is why the current component in the second Ohm’s Law equation is i1 minus i2.
Next these two Ohm’s Law equations are substituted in the original equation which gives the 4th line below.
From the 4th line the rest is just algebra. Then do the exact same process for loop 2 as seen above. This gives 2 equations with 2 unknowns. Of course these can be solved by any linear algebra method you prefer.
Below I solved it using simple substitution. This solves for the unknown currents.
Finally, substitute these found currents back into the Ohm’s Law equations we developed earlier and you find the unknown voltage drops. The last diagram shows the final solution.
I hope this answers my reader’s question and has been helpful. Please let me know if anyone needs more help.