Ohm's Law and Kirchhoff's Laws for Circuit Analysis

Whether you are wiring a UPS backup at home, studying for FSc physics or taking a first course in electrical engineering, almost every DC circuit can be solved with just three rules: Ohm's law, Kirchhoff's current law and Kirchhoff's voltage law. Together they turn a tangle of resistors and batteries into a set of simple linear equations. This article shows how to apply them systematically and how to check that your answers are right.

Ohm's Law

In 1827 the German physicist Georg Simon Ohm showed that, for many conductors at constant temperature, the current through them is proportional to the voltage across them:

$$ V = IR $$

Here \( V \) is the potential difference in volts (V), \( I \) is the current in amperes (A) and \( R \) is the resistance in ohms (\( \Omega \)). The power dissipated as heat in a resistor follows directly:

$$ P = VI = I^2 R = \frac{V^2}{R} $$

Series and Parallel Resistors

Two results save a great deal of time:

• Series (same current through each): \( R_{\text{eq}} = R_1 + R_2 + \cdots \)
• Parallel (same voltage across each): \( \dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots \)

For exactly two resistors in parallel, the shortcut "product over sum" is handy: \( R_{\text{eq}} = R_1R_2/(R_1 + R_2) \).

Kirchhoff's Laws

In 1845, while still a student in Königsberg, Gustav Kirchhoff formulated two laws that extend Ohm's law to whole networks.

Kirchhoff's Current Law (KCL)

The algebraic sum of currents entering any node is zero:

$$ \sum I_{\text{in}} = \sum I_{\text{out}} $$

This is conservation of charge: charge cannot pile up at a junction.

Kirchhoff's Voltage Law (KVL)

The algebraic sum of voltages around any closed loop is zero:

$$ \sum_{\text{loop}} V = 0 $$

This is conservation of energy: a charge going round a loop and returning to its start point must end up with the same potential energy.

Sign Conventions That Prevent Errors

• Moving through a resistor in the direction of current, the voltage drops by \( IR \).
• Moving through a battery from − to +, the voltage rises by its EMF.
• If you guess a current direction and the answer is negative, the current simply flows the other way. You do not need to redo the problem.

Worked Example 1: A Series–Parallel Circuit

A 12 V battery supplies a 4 Ω resistor \( R_1 \) in series with a parallel combination of \( R_2 = 6 \ \Omega \) and \( R_3 = 12 \ \Omega \). Find all currents and the power in each resistor.

Step 1: Combine the parallel pair.

$$ R_{23} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \ \Omega $$

Step 2: Total resistance.

$$ R_{\text{total}} = R_1 + R_{23} = 4 + 4 = 8 \ \Omega $$

Step 3: Supply current (Ohm's law).

$$ I = \frac{12 \ \text{V}}{8 \ \Omega} = 1.5 \ \text{A} $$

Step 4: Voltages.

$$ V_1 = 1.5 \ \text{A} \times 4 \ \Omega = 6 \ \text{V}, \qquad V_{23} = 1.5 \ \text{A} \times 4 \ \Omega = 6 \ \text{V} $$

Step 5: Branch currents.

$$ I_2 = \frac{6 \ \text{V}}{6 \ \Omega} = 1 \ \text{A}, \qquad I_3 = \frac{6 \ \text{V}}{12 \ \Omega} = 0.5 \ \text{A} $$

Step 6: Power.

$$ P_1 = (1.5)^2(4) = 9 \ \text{W}, \quad P_2 = (1)^2(6) = 6 \ \text{W}, \quad P_3 = (0.5)^2(12) = 3 \ \text{W} $$

Sanity checks. KCL: \( I_2 + I_3 = 1 + 0.5 = 1.5 \) A, the supply current. KVL: \( 6 + 6 = 12 \) V. Power balance: the battery delivers \( 12 \times 1.5 = 18 \) W, and \( 9 + 6 + 3 = 18 \) W. Everything agrees.

Worked Example 2: Two Sources Using Nodal Analysis

Series–parallel reduction fails when there is more than one source. Consider a circuit with one top node X and a common ground. Connected to X are:

• a 10 V source in series with \( R_1 = 2 \ \Omega \),
• a 5 V source in series with \( R_2 = 3 \ \Omega \),
• a resistor \( R_3 = 4 \ \Omega \) straight to ground.

Both sources have their negative terminal on ground. Find the voltage at X and every current.

Step 1: Choose the unknown. Let \( V \) be the voltage of node X relative to ground.

Step 2: Write KCL at X, taking every current as leaving the node:

$$ \frac{V - 10}{2} + \frac{V - 5}{3} + \frac{V}{4} = 0 $$

Step 3: Clear fractions by multiplying by 12:

$$ 6(V - 10) + 4(V - 5) + 3V = 0 $$

$$ 6V - 60 + 4V - 20 + 3V = 0 \quad \Rightarrow \quad 13V = 80 \quad \Rightarrow \quad V = \frac{80}{13} \approx 6.15 \ \text{V} $$

Step 4: Branch currents.

$$ I_1 = \frac{10 - 6.15}{2} = \frac{25}{13} \approx 1.92 \ \text{A} \ \ \text{(from the 10 V source into X)} $$

$$ I_2 = \frac{5 - 6.15}{3} = -\frac{5}{13} \approx -0.385 \ \text{A} $$

$$ I_3 = \frac{6.15}{4} = \frac{20}{13} \approx 1.54 \ \text{A} \ \ \text{(down through } R_3) $$

The negative sign on \( I_2 \) means current actually flows into the positive terminal of the 5 V source: the stronger 10 V source is charging it, just as a car alternator charges a battery.

Sanity checks. KCL: \( 25/13 - 5/13 = 20/13 \) A, so current in equals current out. KVL around the loop containing the 10 V source, \( R_1 \) and \( R_3 \):

$$ 10 - 2\left(\frac{25}{13}\right) - 4\left(\frac{20}{13}\right) = 10 - \frac{130}{13} = 0 $$

The node voltage also lies between 5 V and 10 V, which is reasonable for a node pulled by both sources.

Nodal or Mesh Analysis?

• Nodal analysis uses KCL at each node; choose it when there are fewer nodes than loops.
• Mesh analysis uses KVL around each independent loop with assumed mesh currents; choose it when there are fewer loops.

Both produce simultaneous linear equations, which for larger circuits are solved with matrices or Cramer's rule.

Did You Know?

• Ohm's work was initially received coldly in Germany, but in 1841 the Royal Society of London awarded him the Copley Medal.
• Kirchhoff's laws are approximations: at very high frequencies, where wavelengths are comparable to circuit size, engineers must use Maxwell's equations instead.
• A typical incandescent filament has a much lower resistance when cold than when hot, so it is not "ohmic" over its full operating range.

Key Takeaways

• Ohm's law \( V = IR \) relates voltage, current and resistance; power is \( P = VI \).
• KCL (conservation of charge) and KVL (conservation of energy) apply to any lumped circuit.
• Reduce series–parallel networks first; use nodal or mesh analysis when there are multiple sources.
• Negative currents simply mean the true direction is opposite to your guess.
• Always check with KCL, KVL and a power balance.

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