February 10, 2025

DC Circuit Analysis

DC circuit analysis is about finding voltages and currents throughout a circuit powered by a constant voltage source. The tools are Ohm's Law, Kirchhoff's two laws, and the series/parallel rules. This is the foundation — understand these and you can tackle any resistive circuit.

The Analysis Toolkit

  • Ohm's Law: V = IR. Relates voltage, current, and resistance at any component.
  • KVL: Voltages around any loop sum to zero.
  • KCL: Currents into any node sum to zero.
  • Series/parallel rules: Simplify circuits to equivalent resistances.

Analysing a Simple Series Circuit

    +──[R1=1kΩ]──[R2=2kΩ]──+
    |                       |
   [5V]                     |
    |                       |
    +───────────────────────+

Total resistance: R = 1k + 2k = 3kΩ
Current: I = 5V / 3000Ω = 1.67mA (same through both resistors)
V across R1: 1.67mA × 1kΩ = 1.67V
V across R2: 1.67mA × 2kΩ = 3.33V
Check: 1.67 + 3.33 = 5V ✓

Analysing a Parallel Circuit

    +────[R1=6Ω]────+
    |               |
   [12V]   [R2=12Ω] |
    |               |
    +───────────────+

R_parallel = (6 × 12) / (6 + 12) = 72/18 = 4Ω
I_total = 12V / 4Ω = 3A

I through R1 = 12V / 6Ω = 2A
I through R2 = 12V / 12Ω = 1A
Check: 2A + 1A = 3A ✓

Thevenin's Theorem

Any linear circuit seen from two terminals can be replaced by a single voltage source (V_th) in series with a single resistor (R_th). This is invaluable for analysing what happens when you connect a load to a complex circuit.

To find the Thevenin equivalent:
1. V_th = open-circuit voltage at the terminals
2. R_th = resistance seen at the terminals with all sources zeroed
   (voltage sources → short circuit, current sources → open circuit)

Example: 12V source, R1=4Ω series, R2=8Ω in parallel to output terminals

V_th = 12V × 8/(4+8) = 8V    (voltage divider, open circuit)
R_th = 4Ω ∥ 8Ω = (4×8)/(4+8) = 2.67Ω

Now attach any load R_L and easily find V and I:
I_L = V_th / (R_th + R_L)

Norton's Theorem

The dual of Thevenin — replace the circuit with a current source (I_N) in parallel with R_N (same as R_th):

I_N = V_th / R_th

From the example above: I_N = 8V / 2.67Ω = 3A
R_N = 2.67Ω (same as R_th)

Superposition

In circuits with multiple sources, analyse the effect of each source independently (zero the others) then add the results. Each source's contribution is found separately using standard analysis, then currents and voltages are summed algebraically.

Two sources V1 and V2:
1. Zero V2 (replace with short), find I' due to V1 alone
2. Zero V1 (replace with short), find I'' due to V2 alone
3. Total I = I' + I''

Works only for linear circuits (resistors, capacitors, inductors, dependent sources). Doesn't work for power calculations — you can't superimpose power directly.

Practical Notes

Real DC circuits have non-ideal sources — batteries have internal resistance, power supplies have output impedance. For most low-power electronics this doesn't matter much, but for high-current loads (motors, solenoids, transmitters) the voltage drop across source resistance can be significant. Thevenin's theorem is the right tool to account for this.