🔀 Kirchhoff's Laws and Series Parallel Networks

You now have $V = IR$ in your pocket. That is one equation, one resistor. Real circuits have five, ten, thirty elements, and Ohm's law alone will not crack them. Kirchhoff's laws are the two equations that turn any network into solvable algebra. Learn to apply them cleanly and the rest of this degree gets easier.

Learning objectives

By the end of this lesson you can:

  • State Kirchhoff's current law (KCL) and voltage law (KVL) and apply the correct sign convention.
  • Reduce series and parallel resistor groups to a single equivalent resistance.
  • Solve a single-source network for every branch current and element voltage.
  • Check your answer with KVL around a loop and a power balance.

Quick review

From last lesson: $V = IR$ links voltage across an element to the current through it, and $P = VI = I^2R = V^2/R$ gives the power. Two facts from that lesson carry straight into this one: prefix conversions (kΩ, mA) must happen before you substitute, and power dissipated must come out positive.

[!NOTE] Kirchhoff's laws rest on two conservation principles: charge is conserved (KCL) and energy is conserved (KVL). Everything below is just bookkeeping for those two ideas.

Theory

Kirchhoff's current law (KCL)

At any node, the sum of currents entering equals the sum of currents leaving.

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

Equivalently, with leaving currents counted positive, $\sum I = 0$ at every node. ==A node cannot store charge==, so in steady state whatever flows in must flow out. KCL is your tool for splitting current at junctions.

Kirchhoff's voltage law (KVL)

Around any closed loop, the algebraic sum of voltages is zero.

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

Pick a traversal direction first, then be consistent. Crossing a resistor in the direction of assumed current is a drop (write $-IR$); crossing against the current is a rise. Crossing a source from the short bar to the long bar is a rise.

Series networks

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Elements in series share one path, so {{c1::the same current}} flows through each. Resistances add:

$$ R_{eq} = R_1 + R_2 + \cdots + R_n $$

Voltages divide in proportion to resistance (voltage divider):

$$ V_k = V_{total}\cdot\frac{R_k}{R_{eq}} $$

Parallel networks

Elements in parallel share two nodes, so {{c2::the same voltage}} appears across each. Conductances add:

$$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots $$

For two resistors only, use the product-over-sum shortcut: $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$. Current divides inversely with resistance:

$$ I_1 = I_{total}\cdot\frac{R_2}{R_1 + R_2} $$

The smaller resistor always takes the larger share of current.

Property Series Parallel
Shared quantity Current Voltage
Equivalent R $R_1 + R_2$ $\frac{R_1R_2}{R_1+R_2}$
Effect of adding a branch R increases R decreases
Divider rule Voltage divider Current divider

Solving any mixed network

flowchart TD
    A["🔍<br/>Mark series and parallel groups"] --> B["🧮<br/>Reduce inward to one R_eq"]
    B --> C["⚡<br/>Find source current"]
    C --> D["↩️<br/>Work back out: node voltages, branch currents"]
    D --> E["✅<br/>Verify with KVL and power balance"]

Reduce from the inside out, then walk back out. Do not round intermediate values.

Worked examples

Example 1: series. A 12 V source feeds $R_1 = 4,\Omega$ and $R_2 = 6,\Omega$ in series.

  1. $R_{eq} = 4 + 6 = 10,\Omega$
  2. $I = 12/10 = 1.2,\text{A}$ (same through both)
  3. $V_1 = 1.2 \times 4 = 4.8,\text{V}$, $V_2 = 1.2 \times 6 = 7.2,\text{V}$
  4. KVL check: $4.8 + 7.2 = 12,\text{V}$. Power: $12 \times 1.2 = 14.4,\text{W}$ supplied, $1.2^2 \times 10 = 14.4,\text{W}$ dissipated.

Example 2: mixed. A 12 V source feeds $R_1 = 2,\Omega$ in series with the parallel pair $R_2 = 6,\Omega$ and $R_3 = 3,\Omega$.

  1. Parallel section: $6 \parallel 3 = \frac{18}{9} = 2,\Omega$
  2. $R_{eq} = 2 + 2 = 4,\Omega$, so $I_{total} = 12/4 = 3,\text{A}$
  3. Voltage across the parallel pair: $3 \times 2 = 6,\text{V}$
  4. Branch currents: $I_2 = 6/6 = 1,\text{A}$, $I_3 = 6/3 = 2,\text{A}$. KCL check: $1 + 2 = 3,\text{A}$.
  5. $V_{R1} = 3 \times 2 = 6,\text{V}$, and $6 + 6 = 12,\text{V}$ closes the loop.

Guided practice

Work each one on paper before opening the answer. Write the KCL or KVL equation explicitly, every time.

Exercise 1. A 9 V source feeds $2,\Omega$, $3,\Omega$, $4,\Omega$ in series. Find $R_{eq}$, the current, each voltage, and total power.

Click to reveal hint and answer

Hint: series means one current for all three. $R_{eq} = 9,\Omega$, $I = 1,\text{A}$, $V = 2,\text{V}, 3,\text{V}, 4,\text{V}$. Sum equals 9 V. $P = 9,\text{W}$, and $1^2 \times 9 = 9,\text{W}$ confirms it.

Exercise 2. A 24 V source feeds $8,\Omega$ and $24,\Omega$ in parallel. Find $R_{eq}$, source current, and each branch current.

Click to reveal hint and answer

Hint: $8 \parallel 24 = \frac{192}{32}$. Then use the current divider to check. $R_{eq} = 6,\Omega$, $I_{total} = 4,\text{A}$, $I_{8} = 3,\text{A}$, $I_{24} = 1,\text{A}$. Note the smaller resistor carries three times the current.

Exercise 3. A 20 V source feeds $4,\Omega$ in series with the parallel pair $8,\Omega$ and $24,\Omega$. Find all currents and voltages, then verify with a power balance.

Click to reveal hint and answer

Hint: reduce $8 \parallel 24$ first, then work back out from $R_{eq}$. $8 \parallel 24 = 6,\Omega$, $R_{eq} = 10,\Omega$, $I_{total} = 2,\text{A}$. Parallel section voltage: $2 \times 6 = 12,\text{V}$, so $I_8 = 1.5,\text{A}$ and $I_{24} = 0.5,\text{A}$. $V_{4\Omega} = 8,\text{V}$, and $8 + 12 = 20,\text{V}$. Power: $4 \times 4 = 16,\text{W}$, $12^2/8 = 18,\text{W}$, $12^2/24 = 6,\text{W}$, total $40,\text{W}$, matching $20 \times 2,\text{W}$.

Exercise 4 (KCL drill). At a node, 5 A enters through $R_1$, and 2 A and 1.5 A leave through $R_2$ and $R_3$. Find the current in the fourth branch and state its direction.

Click to reveal answer

$5 - 2 - 1.5 = 1.5,\text{A}$ leaving the node.

Expert corner

  • Sign errors cost more marks than algebra. Draw the arrow for every assumed current before writing a single equation.
  • Parallel does not mean equal current. Current splits inversely with resistance. Voltage splits directly.
  • Adding a parallel branch always lowers $R_{eq}$. If your answer went up, you multiplied instead of dividing somewhere.
  • Always run the power balance. Source power supplied must equal the sum of $I^2R$ in every resistor. It catches almost every arithmetic slip in under a minute.
  • Convert prefixes first. $4.7,\text{k}\Omega$ becomes $4700,\Omega$ before substitution, never after.

Three key relations to memorize:

  • KCL: {{c3::$\sum I_{in} = \sum I_{out}$}}
  • KVL: {{c4::$\sum V = 0$ around any closed loop}}
  • Two-resistor parallel: {{c5::$R_{eq} = \dfrac{R_1R_2}{R_1+R_2}$}}
[
  { "front": "State Kirchhoff's current law.", "back": "At any node, the sum of currents entering equals the sum of currents leaving." },
  { "front": "State Kirchhoff's voltage law.", "back": "The algebraic sum of voltages around any closed loop is zero." },
  { "front": "What is conserved in KCL?", "back": "Charge: a node cannot accumulate charge in steady state." },
  { "front": "What is conserved in KVL?", "back": "Energy: a charge returning to its starting node has zero net energy change." },
  { "front": "Equivalent resistance of n resistors in series?", "back": "R_eq = R1 + R2 + ... + Rn." },
  { "front": "Equivalent resistance of n resistors in parallel?", "back": "1/R_eq = 1/R1 + 1/R2 + ... + 1/Rn." },
  { "front": "Product-over-sum shortcut applies when?", "back": "Exactly two resistors in parallel: R_eq = R1R2/(R1+R2)." },
  { "front": "What quantity is shared by elements in series?", "back": "The same current flows through each." },
  { "front": "What quantity is shared by elements in parallel?", "back": "The same voltage appears across each." },
  { "front": "Voltage divider formula?", "back": "V_k = V_total times R_k / R_eq." },
  { "front": "Two-branch current divider formula?", "back": "I1 = I_total times R2/(R1 + R2)." },
  { "front": "Which parallel branch carries more current?", "back": "The one with the smaller resistance." },
  { "front": "Where do you start reducing a mixed network?", "back": "At the innermost series or parallel group, then work outward." },
  { "front": "How do you verify a solved DC network?", "back": "Check KVL around every loop and confirm source power equals total I squared R loss." }
]

Summary and next steps

KCL splits current at nodes, KVL sums voltage around loops, series adds resistance while parallel adds conductance, and mixed networks fall to reduce-inward, work-outward. Exam habit: arrow every current, write the law explicitly, then verify with a power balance.

Next lesson, Thevenin and Norton Equivalents, takes exactly these tools and compresses a whole network into one source plus one resistor, which is how you stop re-solving the same circuit for every load change.