In the lesson we have been able to find the currents in circuits by combining resistances in series and parallel, and using Ohm’s law. This technique can be used for many circuits. However, some circuits are too complicated for that analysis.
To deal with such complicated circuits, we use Kirchhoff’s rules, devised by G. R. Kirchhoff in the mid-nineteenth century. There are two rules, and they are simply convenient applications of the laws of conservation of charge and energy.
Junction rule (KCL) is based on the conservation of electric charge. It states that at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.
$$ I_1=I_2+I_3 $$
Loop rule (KVL) is based on the conservation of energy. It states that the sum of the changes in potential around any closed loop of a circuit must be zero.
$$ \sum V=0 $$
Lets consider the simple circuit
The positive side of the battery, point $e$ is at a high potential compared to point $d$ at the negative side of the battery. We follow the current around the circuit starting at any point. We choose to start at point $d$ and follow a positive test charge completely around this circuit. As we go, we note all changes in potential. When the test charge returns to point $d$, the potential will be the same as when we started (total change in potential around the circuit is zero).
We plot the changes in potential around the circuit; point $d$ is arbitrarily taken as zero. As our positive test charge goes from point $d$, which is the negative or low potential side of the battery, to point $e$, which is the positive terminal (high potential side) of the battery, the potential increases by $12.0$ V. That is,
$$ V_{ed}=+12\text{ V.} $$
When our test charge moves from point $e$ to point $a$, there is no change in potential since there is no source of emf and we assume negligible resistance in the connecting wires. Next, as the charge passes through the $400\Omega$ resistor to get to point $b$, there is a decrease in potential of
$$ V=IR=(0.0174)(400)=7\text{ V.} $$
The positive test charge is flowing since it is heading toward the negative terminal of the battery, as indicated in the graph. Because this is a decrease in potential, we use a negative sign:
$$ V_{ba}=V_b-V_a=-7\text{ V.} $$
As the charge proceeds from $b$ to $c$ there is another potential decrease (a “voltage drop”) of