Kirchhoff's Laws and Circuit Analysis

Kirchhoff's laws are simply two bookkeeping rules for circuits: charge doesn't vanish, and you can't gain height by walking in a circle. They're named after Gustav Kirchhoff, who stated them in 1845, and you'll often see them shortened to KCL and KVL (Kirchhoff's current law and voltage law). They're what let you find every current in a circuit that series and parallel rules can't untangle. The junction rule $\sum I_{\text{in}} = \sum I_{\text{out}}$ just means (whatever current flows into a junction) equals (whatever flows out of it). The loop rule $\sum \Delta V = 0$ means (add up every voltage rise and drop around a loop) and you get (zero, because you're back where you started). Big circuits look intimidating, but they're just these two rules used over and over. Bear with it, and the algebra does the rest.

Which way does the current go? Jump-start a car and three things end up wired together: a flat battery, a good battery and the starter motor. Series and parallel rules can't untangle that, and you can't even guess every current's direction in advance. So what can we do? It turns out two conservation laws are enough, every time.

The junction rule (KCL). In a steady circuit charge can't pile up anywhere, so at every junction the currents flowing in add up to the currents flowing out. You can say the same thing with the current density from last lesson: the flux of $\mathbf{J}$ out of a small closed surface around the junction is $-dQ_{\text{inside}}/dt = 0$. It's the current version of Gauss's law.

The loop rule (KVL). In a steady circuit the electric field is electrostatic, and electrostatic fields are conservative: $\oint\mathbf{E}\cdot d\mathbf{l} = 0$ around any closed path. Since $\Delta V = -\int\mathbf{E}\cdot d\mathbf{l}$, the potential changes around any loop add to zero. Think of hiking a loop on a mountain. However much you climb and descend, you end up back at the altitude you started at. Walk a loop in a circuit and you end up back at the potential you started at.

Sign rules for walking a loop:

• Across a resistor in the direction of its current: $-IR$ (current flows downhill in potential). Against the current: $+IR$.
• Across a battery from its $-$ terminal to its $+$ terminal: $+\mathcal{E}$. From $+$ to $-$: $-\mathcal{E}$.

Series and parallel, derived. You've probably memorized these rules already. Notice they're really just KCL and KVL in disguise. Resistors in series carry the same current (KCL: it has nowhere else to go), and their drops add (KVL): $V = IR_1 + IR_2$, so $R_s = R_1 + R_2$. Resistors in parallel share one voltage (KVL around the small loop they form), and their currents add (KCL): $I = V/R_1 + V/R_2$, so $1/R_p = 1/R_1 + 1/R_2$. Two resistors in series also split a voltage in proportion. That's the voltage divider: $V_2 = \frac{R_2}{R_1 + R_2}V$.

Real batteries. A battery's chemistry supplies an EMF $\mathcal{E}$, but the current also has to push through the battery's own internal resistance $r$. So the terminal voltage sags under load: $V = \mathcal{E} - Ir$.

The method.

1. Give every branch a current with a guessed direction.
2. Write KCL at the junctions. With $j$ junctions only $j - 1$ of these equations are independent; the last one repeats the others.
3. Write KVL around enough loops to match the number of unknowns, with consistent signs.
4. Solve the linear system. A negative answer just means that current flows opposite to your guess, and nothing needs redoing.

Circuits are linear systems. Take the circuit drawn in the first picture further down: a 12 V source with $R_1 = 2\ \text{k}\Omega$ on the left, a $4\ \text{k}\Omega$ load in the middle, and a second source $V_2$ with $R_2 = 2\ \text{k}\Omega$ on the right. Call $I_1$ the current up the left branch, $I_3$ the current down the load and $I_2$ the current down the right branch. With resistances in kΩ and currents in mA, every $IR$ comes out in volts. KCL at the top junction and KVL around the left and right loops give

Row reduction (Systems of Equations) gives $I_1 = 3.6 - 0.2V_2$, $I_2 = 2.4 - 0.3V_2$ and $I_3 = 1.2 + 0.1V_2$. Notice every current is linear in the source voltages. That's superposition, and it makes sense that it holds for any circuit of resistors and sources: the equations are linear, so their solutions are too.

A shortcut: mesh currents. Give each loop its own circulating current, $J_1$ in the left loop and $J_2$ in the right, both clockwise. A branch shared by two loops carries the difference. Think about what that buys you: whatever circulates into a junction also circulates out, so KCL holds automatically and only the loop equations remain:

The matrix is symmetric. Each loop's total resistance sits on the diagonal, and the shared $4\ \text{k}\Omega$ resistor appears off the diagonal with a minus sign. Each equation is a line in the $(J_1, J_2)$ plane, and the second picture further down shows the currents as the point where the lines cross. The branch currents follow: $I_1 = J_1$, $I_2 = J_2$, $I_3 = J_1 - J_2$.

A common mistake. When a current comes out negative, it's tempting to think you made an error and go back to flip signs in the middle of the solution. Don't. A negative current just means it flows opposite to your arrow, and the rest of the algebra is still right. Two other slips cause most wrong answers: switching sign conventions halfway around a loop, and writing one KCL equation too many, which makes the matrix singular.

KVL rests on $\oint\mathbf{E}\cdot d\mathbf{l} = 0$. In Faraday's Law we'll see a changing magnetic flux break that, and inductors get a voltage term of their own.

Circuit simulators used to design every chip and circuit board run this lesson at scale: they write KCL at every node and solve sparse linear systems with millions of unknowns. Power engineers solve the same equations for a whole grid to see how current redistributes when a line trips, and Wheatstone bridges turn tiny resistance changes into readable voltages in scales, pressure sensors and strain gauges.