DC Circuit Analysis

DC circuit analysis is about finding voltages and currents throughout a circuit powered by a constant (direct current) voltage source. The complete toolkit is small: Ohm's Law, Kirchhoff's two laws, and the series/parallel combination rules. These four tools, used methodically, can solve any purely resistive DC circuit. The harder question is which method to reach for first — this covers all of them.

The Analysis Toolkit

ToolWhat it saysBest used when
Ohm's Law (V=IR)Voltage, current, and resistance at any single componentSingle-element calculations, once I or V is known
KVLVoltages around any closed loop sum to zeroMesh analysis — multiple loops with unknown currents
KCLCurrents into any node sum to zeroNodal analysis — multiple nodes with unknown voltages
Series/parallel reductionSimplify groups of resistors to one equivalentCircuits that can be collapsed step-by-step
Thevenin's theoremAny two-terminal circuit = V_th + R_th in seriesAnalysing the effect of a changing load
Norton's theoremAny two-terminal circuit = I_N in parallel with R_NParallel current source representation
SuperpositionLinear circuits with multiple sources: solve each independently, then addMulti-source circuits, checking individual contributions

Simple Series Circuit

Simple Series Circuit A closed loop with a 5V supply on the left and two resistors, R1 (1k ohms) and R2 (2k ohms), in series on the top branch. 5V R1=1kΩ R2=2kΩ
Two resistors in series, forcing the same current to flow through both.
Total R = 1k + 2k = 3kΩ
I = 5V / 3000Ω = 1.667 mA (same through both)

V_R1 = 1.667 mA × 1kΩ = 1.667V
V_R2 = 1.667 mA × 2kΩ = 3.333V
KVL check: 1.667 + 3.333 = 5V ✓

Simple Parallel Circuit

Simple Parallel Circuit A circuit with a 12V supply on the left. The top and bottom wires extend to the right, with R1 (6 ohms) and R2 (12 ohms) connected in parallel across them. 12V R1=6Ω R2=12Ω
Two resistors in parallel, experiencing the same voltage drop across them.
R_parallel = (6 × 12) / (6 + 12) = 4Ω
I_total = 12V / 4Ω = 3A

I_R1 = 12V / 6Ω = 2A
I_R2 = 12V / 12Ω = 1A
KCL check: 2 + 1 = 3A ✓

Thevenin's Theorem

Any linear two-terminal circuit can be replaced by a single voltage source (V_th) in series with a single resistor (R_th). This simplification is enormously useful when you want to analyse the effect of connecting different loads to a circuit — instead of redoing the full analysis for every load, you find the Thevenin equivalent once, then V and I for any load is trivial.

Procedure:
  1. Remove the load from the two terminals
  2. V_th = open-circuit voltage at those terminals
  3. Zero all independent sources (V sources → short, I sources → open)
  4. R_th = resistance seen looking into the terminals

Then with any load R_L:
  I_L = V_th / (R_th + R_L)
  V_L = I_L × R_L = V_th × R_L / (R_th + R_L)

Thevenin Example

Thevenin Equivalent Circuit A 12V supply on the left connects to R1 (4 ohms) in series, which then meets a node. From the node, R2 (8 ohms) goes to ground, and terminal A goes out to the right. Terminal B is ground. 12V R1=4Ω R2=8Ω A B
Finding the Thevenin equivalent of a linear circuit at terminals A and B.
Step 1 — V_th (open circuit, no load between A and B):
  Voltage divider: V_th = 12V × 8/(4+8) = 8V

Step 2 — R_th (zero the 12V source = short circuit):
  R_th = R1 ∥ R2 = (4 × 8)/(4 + 8) = 2.67Ω

Thevenin equivalent: 8V source in series with 2.67Ω

With a 10Ω load:
  I_L = 8V / (2.67 + 10) = 0.631A
  V_L = 0.631 × 10 = 6.31V

With a 1Ω load:
  I_L = 8V / (2.67 + 1) = 2.18A
  V_L = 2.18V

Norton's Theorem

Norton equivalent = current source I_N in parallel with R_N

I_N = V_th / R_th   (or short-circuit current at the terminals)
R_N = R_th          (same resistance)

From the example above: I_N = 8V / 2.67Ω = 3A, R_N = 2.67Ω

Thevenin ↔ Norton conversions:
  V_th = I_N × R_N
  I_N = V_th / R_th
  R_N = R_th (identical)

Superposition

For circuits with multiple independent sources: analyse the contribution of each source separately, zero the others, then sum the results. Works for voltage and current but not directly for power (power is non-linear).

Circuit: two voltage sources V1 and V2 with resistors R1, R2, R3

Step 1: Zero V2 (replace with short wire). Find I' due to V1 alone.
Step 2: Zero V1 (replace with short wire). Find I'' due to V2 alone.
Step 3: I_total = I' + I''  (algebraic sum — watch directions)

Superposition Example

Superposition Circuit A T-network. V1 (12V) on the left goes through R1 (4 ohms) to a central node. V2 (6V) on the right goes through R2 (6 ohms) to the same central node. R3 (12 ohms) goes from the central node to ground. V1=12V R1=4Ω R2=6Ω V2=6V R3=12Ω GND
A multi-source circuit analyzed using superposition.
Contribution of V1 alone (V2 shorted):
  R2 ∥ R3 = (6×12)/(6+12) = 4Ω
  R_total = R1 + 4 = 8Ω
  I1' = 12/8 = 1.5A through R1
  V_node' = 1.5 × 4 = 6V
  I_R3' = 6/12 = 0.5A (downward)

Contribution of V2 alone (V1 shorted):
  R1 ∥ R3 = (4×12)/(4+12) = 3Ω
  R_total = R2 + 3 = 9Ω
  I2'' = 6/9 = 0.667A through R2
  V_node'' = 0.667 × 3 = 2V
  I_R3'' = 2/12 = 0.167A (downward)

Total current through R3: I_R3 = 0.5 + 0.167 = 0.667A
Total node voltage: 6 + 2 = 8V

Source Transformations

A convenient shortcut that converts between Thevenin (voltage source + series resistor) and Norton (current source + parallel resistor) forms:

FromToTransformation
Thevenin (V_th, R_th)NortonI_N = V_th / R_th, same R in parallel
Norton (I_N, R_N)TheveninV_th = I_N × R_N, same R in series

Real Sources

Ideal voltage sources don't exist — batteries and power supplies have internal resistance. A real battery is its open-circuit voltage in series with an internal resistance r:

V_terminal = V_open_circuit − I × r

Battery: V_oc = 12V, r = 0.5Ω
Load = 10Ω:
  I = 12 / (0.5 + 10) = 1.143A
  V_terminal = 12 − 1.143 × 0.5 = 11.43V

Under heavy load (1Ω):
  I = 12 / (0.5 + 1) = 8A
  V_terminal = 12 − 8 × 0.5 = 8V  ← significant sag

References