Comparison

Thevenin vs Norton Theorem

Before connecting a load to a complex amplifier output stage, a circuit designer reduces the entire amplifier to a single voltage source and series resistance — that is Thevenin's theorem at work. The same network can equally be represented as a current source in parallel with a resistance — Norton's form. Choosing the right representation before solving cuts calculation time in half, especially in transistor bias and filter design problems.

EEE, ECE, EI

Side-by-side comparison

ParameterTheveninNorton Theorem
Equivalent CircuitV_th in series with R_thI_N in parallel with R_N
Source TypeVoltage source (V_th)Current source (I_N)
Equivalent ResistanceR_th (same calculation as R_N)R_N = R_th (identical value)
RelationshipV_th = I_N × R_NI_N = V_th / R_th
Best for LoadHigh-impedance loads; voltage divider analysisLow-impedance loads; current divider analysis
Finding Open-Circuit ValueMeasure/calculate V_oc at terminalsMeasure/calculate I_sc at terminals
Preferred in Transistor CircuitsCommon-emitter voltage bias networksCurrent mirror and current source analysis
Source ConversionConvert to Norton: I_N = V_th / R_thConvert to Thevenin: V_th = I_N × R_N

Key differences

Thevenin and Norton equivalents represent the same network — R_th and R_N are always equal. The difference is source form: Thevenin uses V_oc as a series voltage source, Norton uses I_sc = V_oc / R_th as a parallel current source. For a 12 V source with 600 Ω internal resistance, the Thevenin form is 12 V + 600 Ω series; the Norton form is 20 mA || 600 Ω. Thevenin suits high-load-impedance calculations; Norton suits current-divider chains and op-amp summing junctions.

When to use Thevenin

Use Thevenin's theorem when the load is a high-impedance element like an op-amp input or a voltmeter — for example, finding the voltage across a 10 kΩ load connected to a bridge sensor network.

When to use Norton Theorem

Use Norton's theorem when the load draws significant current or when current dividers dominate — for example, analyzing a BJT current mirror where the collector current is found by current division from I_N.

Recommendation

For most university exam problems, choose Thevenin — question setters ask for "the voltage across the load," which maps directly to the Thevenin equivalent with a voltage divider finish. Switch to Norton only when the problem explicitly involves current sources or current dividers.

Exam tip: The key examiner trap is asking you to find R_th with dependent sources present — you must apply a test voltage V_x, find the resulting I_x, and compute R_th = V_x / I_x; setting independent sources to zero and using ohmmeter rules fails with dependent sources.

Interview tip: A placement interviewer at L&T or ABB will ask you to explain source transformation in one line — say: "V_th and I_N are interchangeable via I_N = V_th / R_th; the parallel or series resistance is the same R_th either way."

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