Circuit Theory and Kirchhoff’s Laws | Digital Electronics

Kirchhoff’s Current Law (KCL)

Kirchhoff’s Current Law states: the algebraic sum of currents entering any node (junction) is zero. Equivalently, the total current flowing into a node equals the total current flowing out of that node.

For a node with three branches carrying currents I₁, I₂, I₃ (with convention that currents entering are positive, leaving are negative):

$I_1 + I_2 - I_3 = 0 \quad\text{or}\quad I_1 + I_2 = I_3$

KCL is a statement of conservation of charge: charge cannot accumulate at a point, so whatever flows in must flow out.

Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law states: the algebraic sum of all voltages around any closed loop in a circuit is zero. When traversing a loop, voltage drops across components and voltage rises across sources must sum to zero.

For a loop containing a voltage source V₁ and resistors with drops V₂, V₃, V₄, V₅, and another source V₆ (with polarity opposite to traversal):

$V_1 - V_2 - V_3 - V_4 - V_5 + V_6 = 0$

KVL follows from conservation of energy: the net energy per coulomb around a closed path must be zero.

Node voltage analysis: systematic approach

Node voltage analysis is a systematic method for solving resistor networks:

Choose a reference node (ground, 0V).
Label the voltages at all other nodes relative to ground.
Apply KCL at each non-reference node, expressing branch currents in terms of node voltages and resistances using Ohm’s Law.
Solve the resulting system of linear equations.

Examples Paper 3, Question 4

For the circuit shown, (a) What is the current through the 1 Ω resistor? (b) What is voltage V₁? (c) What power is dissipated in each of the 4 Ω resistors?

The circuit has a 6V source, a 1 Ω series resistor, and then two 4 Ω resistors in parallel. The parallel combination is connected between the 1 Ω resistor and ground.

Step 1: Simplify the parallel resistors.

The two 4 Ω resistors are in parallel:

$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2\;\Omega$

Step 2: Reduce to a single loop.

The 1 Ω resistor is in series with the 2 Ω equivalent, giving total resistance:

$R_{\text{total}} = 1 + 2 = 3\;\Omega$

Step 3: Find total current.

From Ohm’s Law across the whole circuit:

$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{6}{3} = 2\text{ A}$

This is the current through the 1 Ω resistor. (Answer (a): 2 A.)

Step 4: Find V₁.

V₁ is the voltage across the parallel combination, i.e., the voltage at the node between the 1 Ω resistor and the two 4 Ω resistors. Using Ohm’s Law on the 2 Ω equivalent:

$V_1 = I_{\text{total}} \times R_{\text{parallel}} = 2 \times 2 = 4\text{ V}$

Or equivalently, V₁ = 6V - (2A × 1Ω) = 4V. (Answer (b): 4V.)

Step 5: Power in each 4 Ω resistor.

The 4V is across each 4 Ω resistor. For each:

$P = \frac{V^2}{R} = \frac{4^2}{4} = \frac{16}{4} = 4\text{ W}$

(Answer (c): 4W per 4 Ω resistor.)

Examples Paper 3, Question 5

For the circuit with a 6V source and a 10V source, find (a) the current through the 20 Ω resistor, (b) the voltages V_AB, V_AC, and V_AD.

The circuit: between nodes A (ground) and D there is a 10V source. From A to B there is a 1 Ω resistor in series with a 6V source (note: the 6V source’s negative terminal is at A). Node B connects to C through a 20 Ω resistor. From C to D there is a 2 Ω resistor. Also: from B to ground (A) there is a 2 Ω resistor, and from C to ground (A) there is a 2 Ω resistor, and from D to ground (A) is the 10V source.

Let us restate this clearly: node A is 0V (ground). Between A and B: 6V source (with the negative at A, so V_B = 6V + drop across 1Ω) plus a 1 Ω resistor in series. Between B and C: 20 Ω resistor. Between C and D: 2 Ω resistor. Between D and A: 10V source (with negative at A, positive at D, so V_D = 10V). Also: a 2 Ω resistor from B to A, and a 2 Ω resistor from C to A.

Node voltage approach. Let V_A = 0V. Known: V_D = 10V. Unknown: V_B, V_C.

Apply KCL at node B. Currents leaving B: through 1 Ω to A (but note the 6V source in series: the voltage across the 1 Ω is V_B - 6V, since the source lifts the voltage); through 2 Ω to A: (V_B - 0)/2; through 20 Ω to C: (V_B - V_C)/20.

$\frac{V_B - 6}{1} + \frac{V_B - 0}{2} + \frac{V_B - V_C}{20} = 0$

Multiply by 20:

$20(V_B - 6) + 10V_B + (V_B - V_C) = 0$

$20V_B - 120 + 10V_B + V_B - V_C = 0$

$31V_B - V_C = 120 \quad\text{(1)}$

Apply KCL at node C. Currents leaving C: through 2 Ω to A: V_C/2; through 2 Ω to D: (V_C - 10)/2; through 20 Ω to B: (V_C - V_B)/20.

$\frac{V_C}{2} + \frac{V_C - 10}{2} + \frac{V_C - V_B}{20} = 0$

Multiply by 20:

$10V_C + 10(V_C - 10) + (V_C - V_B) = 0$

$10V_C + 10V_C - 100 + V_C - V_B = 0$

$21V_C - V_B = 100 \quad\text{(2)}$

From (2): V_B = 21V_C - 100. Substitute into (1):

$31(21V_C - 100) - V_C = 120$

$651V_C - 3100 - V_C = 120$

$650V_C = 3220$

$V_C = \frac{3220}{650} = 4.95\text{ V}$

Then V_B = 21(4.95) - 100 = 104.00 - 100 = 4.00 V (approximately). Exactly:

$V_B = 21 \times \frac{3220}{650} - 100 = \frac{67620}{650} - 100 = 104.03 - 100 = 4.03\text{ V}$

Let us recompute exactly: V_C = 3220/650 = 322/65 ≈ 4.9538 V. V_B = 21 × 322/65 - 100 = 6762/65 - 100 = (6762 - 6500)/65 = 262/65 = 4.0308 V.

Current through the 20 Ω resistor:

$I_{20} = \frac{V_B - V_C}{20} = \frac{262/65 - 322/65}{20} = \frac{-60/65}{20} = \frac{-60}{1300} = -\frac{6}{130} = -\frac{3}{65} \approx -0.0462\text{ A}$

The negative sign means current flows from C to B through the 20 Ω resistor. Magnitude: 46.2 mA. (Answer (a).)

Node voltages: V_AB = V_B - V_A = 4.03V, V_AC = V_C - V_A = 4.95V, V_AD = V_D - V_A = 10V. (Answer (b).)

Series-parallel reduction technique

The reduction approach used in Question 4 works well when the circuit has a clear series-parallel structure:

Identify purely parallel branches and replace with their equivalent resistance.
Identify series chains and replace with their sum.
Repeat until the circuit is reduced to a single loop.
Apply Ohm’s Law to find the total current.
Expand back step by step, using voltage division and current division to recover individual branch values.

This technique can be quicker than formal node analysis for circuits with a simple hierarchical structure. For circuits with multiple sources and cross-connections (like Question 5), node voltage analysis or mesh current analysis is more systematic.

Tripos relevance

Resistor network analysis questions appear on Paper 2, typically as part of a larger question that combines electronics with Boolean logic or processor architecture. The examples from Examples Paper 3 (Questions 4 and 5) are representative of the difficulty expected. You should be fluent in both series-parallel reduction and node voltage analysis, and you should work through these questions until the methods are second nature.

See also /modules/digital-electronics/11-electronics-fundamentals/03-potential-dividers/ for the potential divider, which is a special case at the heart of transistor circuit analysis.