Ohm's Law, Kirchhoff's Laws and the Division Rules

By the end you should be able to: State Ohm's law and Kirchhoff's current and voltage laws, apply the voltage and current division rules, and calculate the potential of any node relative to a chosen reference.

Everything in this module rests on three laws. They are simple enough to state in a line each, and almost every circuit error in the exam comes from applying them with the wrong sign.

Ohm's law

V = IR

Valid for a linear, ohmic element at constant temperature. A diode does not obey it — Module 3 shows why — and neither does a filament lamp once it heats.

Kirchhoff's current law (KCL)

The sum of currents entering a node equals the sum leaving.

\sum I_{in} = \sum I_{out} \qquad\text{or}\qquad \sum I = 0 \text{ at a node}

It is conservation of charge. Charge does not pile up at a junction, so whatever arrives must depart.

Kirchhoff's voltage law (KVL)

Around any closed loop, the algebraic sum of EMFs and potential drops is zero.

\sum V = 0 \text{ around a loop}

It is conservation of energy: carry a unit charge once round a loop and it returns to where it started, so the net work is zero.

The sign convention, which is where the marks are lost. Pick a direction to traverse the loop and stay with it:

Through a resistor in the direction of assumed current, the potential falls: enter as $-IR$ .
Through a resistor against the assumed current: $+IR$ .
Through a source from − to +: $+E$ (a rise).
Through a source from + to −: $-E$ .

If you assumed a current backwards, the algebra returns a negative number and the magnitude is still right. Do not go back and "fix" the arrow — the sign is the answer telling you the true direction.

Voltage division

For resistors in series the current is common, so voltage divides in proportion to resistance:

V_1 = V \cdot \frac{R_1}{R_1 + R_2 + \cdots}

The largest resistor takes the largest share.

Current division

For resistors in parallel the voltage is common, so current divides in inverse proportion:

I_1 = I \cdot \frac{R_2}{R_1 + R_2} \qquad\text{(two resistors only)}

Note the numerator is the other resistor. That inversion is the most common slip in the whole topic. Sanity-check every answer: more current must go through the smaller resistance.

For more than two branches, work with conductances $G = 1/R$ :

I_k = I \cdot \frac{G_k}{\sum G}

Relative potential

Potential has no absolute value; only differences are physical. So we choose a reference node — the datum, ground, or 0 V — and quote every other node relative to it.

V_{AB} = V_A - V_B

Two consequences worth being clear about:

Moving the reference changes every node potential but no voltage difference, and therefore no current. The circuit does not know where you put your zero.
$V_{AB} = -V_{BA}$ . Getting a negative answer means the node is below the reference, not that you made a mistake.

This is why a bird on a high-voltage line is unharmed: it sits at several hundred kilovolts relative to earth, but both feet are at nearly the same potential, and it is the difference across the bird that would drive current.