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Electrical Circuits and Sensors · Module 2 of 10

Kirchhoff's Laws and Resistive Networks

Ohm's law handles one resistor. Real circuits have many, joined at nodes and arranged in loops. Two conservation laws, one for voltage and one for current, let you solve any of them.

01

Readiness check

This module builds on Ohm's law. Tick only what you can do closed-notes.

  • Apply V = IR to a single resistor.
  • Tell a series connection from a parallel one.
  • Add fractions with different denominators.
  • Recognise a node and a loop in a schematic.
  • Recall that energy and charge are conserved.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit Ohm's law and power in Module 1.
3 or more weak itemsReview charge conservation in Physics: Electricity and Magnetism.
02

The core idea

Charge is conserved at every node, and energy is conserved around every loop. Kirchhoff's two laws turn those facts into equations that solve any resistive network.

KCL: Σ Iin = Σ IoutKVL: Σ Vrises = Σ Vdropsseries: Req = R1 + R2 + …

Kirchhoff's current law (KCL) says the currents entering a node must equal the currents leaving it, because charge cannot pile up. Kirchhoff's voltage law (KVL) says the voltage rises and drops around any closed loop must cancel, because a charge returning to where it started has gained no net energy. From these two laws come the familiar shortcuts: resistors in series share the same current and add directly, resistors in parallel share the same voltage and add as reciprocals, and the divider rules read off a voltage or current without solving the whole circuit. Every analysis method in the next module is just an organised way of writing KCL and KVL.

The skill works when: you pick a consistent sign convention and apply the matching law at each node or loop.
The skill breaks down when: series and parallel are mixed up, or current directions are assigned inconsistently.
The concept. Around a loop the source voltage equals the sum of the drops (KVL). At a node the current in equals the sum of the currents out (KCL). The two laws together close any network.
03

The skills, taught in order

Five skills move from the two laws to the series and parallel shortcuts and the divider rules that follow.

2.1 Kirchhoff's current law

At any node the algebraic sum of currents is zero: what flows in must flow out. Assign each branch a current with an assumed direction; a negative answer simply means the real current flows the other way. KCL expresses conservation of charge.

2.2 Kirchhoff's voltage law

Around any closed loop the algebraic sum of voltage changes is zero. Pick a direction to travel the loop, count a rise across a source and a drop across a resistor, and set the total to zero. KVL expresses conservation of energy.

2.3 Series resistance

Resistors in series carry the same current. Their voltages add, so the equivalent resistance is the sum: Req = R1 + R2 + … A series chain always has a larger resistance than any one element.

2.4 Parallel resistance

Resistors in parallel share the same voltage. Their currents add, so the reciprocals add: 1/Req = 1/R1 + 1/R2 + … A parallel combination is always smaller than the smallest element. For two, Req = R1R2/(R1 + R2).

ConnectionShared quantityEquivalent resistanceDivider rule
SeriescurrentR1 + R2Vk = V · Rk/Req
ParallelvoltageR1R2/(R1 + R2)Ik = I · Req/Rk

The two basic connections, what they share, and the divider that reads off a single voltage or current.

2.5 Voltage and current dividers

In a series string the voltage splits in proportion to each resistance: Vk = V · Rk/Req. In a parallel pair the current splits in inverse proportion to each resistance. These dividers are the workhorses of sensor and reference circuits.

Engineering connection: a potentiometer position sensor is a voltage divider, and reading its output is exactly the divider rule applied to a moving tap.

04

Worked example 1: a series loop by KVL

A 20 V source drives two resistors in series, R1 = 8 Ω and R2 = 12 Ω. Find the loop current and the voltage across R2.

Figure 1. The two series resistors carry one common current. KVL sets the source voltage equal to the two drops, and the divider gives the drop across R2.
  1. ProblemFind the loop current and the voltage across R2 in Figure 1.
  2. Given / findV = 20 V, R1 = 8 Ω, R2 = 12 Ω. Find I and V2.
  3. AssumptionsIdeal source; series connection, so one current flows.
  4. ModelAdd the series resistances, find the current by Ohm's law, then the drop across R2.
  5. EquationsReq = R1 + R2I = V / ReqV2 = I R2
  6. SolveReq = 8 + 12 = 20 Ω. I = 20 / 20 = 1 A. V2 = 1 × 12 = 12 V.
  7. CheckKVL: V1 + V2 = (1 × 8) + 12 = 8 + 12 = 20 V, equal to the source. The divider agrees: V2 = 20 × 12/20 = 12 V.
  8. ConclusionThe 1 A current is common to both resistors, and the larger resistor takes the larger share of the voltage.
Result. I = 1 A and V2 = 12 V.
05

Worked example 2: parallel branches by KCL

A 24 V source feeds three resistors in parallel: R1 = 6 Ω, R2 = 12 Ω, R3 = 4 Ω. Find the current in each branch and the total current from the source.

Figure 2. Each parallel branch sees the full 24 V, so its current is 24/R. The branch currents add by KCL to give the total drawn from the source.
  1. ProblemFind the three branch currents and the total current in Figure 2.
  2. Given / findV = 24 V across R1 = 6 Ω, R2 = 12 Ω, R3 = 4 Ω. Find I1, I2, I3, I.
  3. AssumptionsIdeal source; all three branches see the same 24 V.
  4. ModelApply Ohm's law to each branch, then KCL to sum the branch currents.
  5. EquationsIk = V / RkI = I1 + I2 + I3
  6. SolveI1 = 24/6 = 4 A. I2 = 24/12 = 2 A. I3 = 24/4 = 6 A. I = 4 + 2 + 6 = 12 A.
  7. CheckThe parallel equivalent is 1/Req = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12, so Req = 2 Ω, and I = 24/2 = 12 A, which matches.
  8. ConclusionThe smallest resistor carries the most current, and the branch currents sum to the source current, as KCL requires.
Result. I1 = 4 A, I2 = 2 A, I3 = 6 A, total I = 12 A.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Adding parallel resistors directlyEquivalent larger than an element"Is the result smaller than the smallest R?"Parallel resistances add as reciprocals.
Inconsistent current directionsKCL never balances"Did I keep one sign convention?"Fix a direction per branch and stay with it.
Same current in parallelEqual currents assumed in unequal branches"Do these share voltage or current?"Parallel branches share voltage, not current.
Divider on the wrong resistorVoltage assigned to the small R"Which R is the drop across?"Vk = V · Rk/Req uses that branch's own Rk.
07

Practice ladder

Level 1 · Direct skill

Two resistors, 10 Ω and 40 Ω, are in series across 25 V. Find the current.

Show answer

Req = 50 Ω, I = 25/50 = 0.5 A.

Level 2 · Mixed concept

For the same series pair, find the voltage across the 40 Ω resistor using the divider rule.

Show answer

V = 25 × 40/50 = 20 V. Check: the 10 Ω takes 5 V, and 20 + 5 = 25 V.

Level 3 · Independent problem

A 30 Ω and a 60 Ω resistor are in parallel across 18 V. Find each branch current, the total current, and the equivalent resistance.

Show answer

I1 = 18/30 = 0.6 A, I2 = 18/60 = 0.3 A, total 0.9 A. Req = (30 × 60)/(30 + 60) = 1800/90 = 20 Ω, and 18/20 = 0.9 A confirms it.

Transfer task | Real engineering

A potentiometer divider must drop a 5 V supply to 2 V at its tap. Choose two resistances that achieve it and state the current they draw.

What good work looks like

Any pair with Rtap/Rtotal = 2/5, for example 2 kΩ and 3 kΩ (tap across the 2 kΩ). The current is 5/(5 kΩ) = 1 mA, a sensible divider load.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my branch currents satisfy KCL at this node."
"Give me a mixed series-parallel network; I will reduce it step by step."
"Solve this whole circuit." Writing the node and loop equations is the skill.
"Is this series or parallel?" Reading the topology yourself is the point.

Portfolio task

Take a real two-resistor sensor divider or a parallel load bank, reduce it to an equivalent resistance, and confirm your branch currents add up by KCL.

Must include: the topology, an equivalent resistance, the branch currents, and a KCL or KVL check.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State KCL and KVL.

KCL: currents into a node equal currents out. KVL: voltage changes around a loop sum to zero.

2. How do series resistances combine?

They add directly: Req = R1 + R2 + …

3. How do parallel resistances combine?

Their reciprocals add; for two, Req = R1R2/(R1 + R2).

4. What does each connection share?

Series shares current; parallel shares voltage.

5. Write the voltage divider rule.

Vk = V · Rk/Req for a resistor in a series string.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the series and parallel results from a blank page.
+3 daysReduce three new mixed networks.
+7 daysCarry KCL and KVL into nodal and mesh analysis, Module 3.
+30 daysReuse the divider when reading sensor outputs.
10

Textbook mapping

This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.

Topic in this moduleWhere to read more
Kirchhoff's current and voltage lawsAlexander & Sadiku, Chapter 2
Series and parallel resistorsAlexander & Sadiku, Chapter 2
Voltage and current dividersAlexander & Sadiku, Chapter 2

Chapter numbers refer to the 4th edition. The laws are standard, so any recent edition will align closely.