Which pairing correctly shows the duality between Thevenin and Norton equivalents?

Voltage source in series with R_th current source in parallel with R_N

Current source with R_th in series voltage source with R_N in parallel

Voltage source with R_th voltage source with R_N

Voltage source in parallel with R_th current source in series with R_N

In the Thevenin and Norton equivalents, what do V_th and I_N represent?

V_th is the open-circuit voltage and I_N the short-circuit current

Both are the load-current values

Both are the open-circuit values

V_th is the short-circuit voltage and I_N the open-circuit current

Which set of relations links the two equivalents?

I_N = V_th/R_th, V_th = I_N R_N and R_N = R_th

I_N = V_th + R_th and R_N = R_th/2

I_N = R_th/V_th and R_N = 2 R_th

I_N = V_th R_th and R_N = 1/R_th

A 12 V source with 4 ohm in series feeds a 6 ohm load. What is the Norton equivalent and the load current?

I_N = 3 A, R_N = 4 ohm, I_L = 1.2 A

I_N = 3 A, R_N = 6 ohm, I_L = 1.5 A

I_N = 12 A, R_N = 4 ohm, I_L = 3 A

I_N = 2 A, R_N = 4 ohm, I_L = 0.8 A

2070 Asar · 6 marks

Thevenin and Norton duality

Short answer

Thevenin’s series voltage source with Rth is dual to Norton’s parallel current source with RN, where Vth = IN × RN and Rth = RN. For a 12 V, 4 Ω network, the Norton equivalent is 3 A in parallel with 4 Ω.

"Thevenin's theorem and Norton's theorem are dual of each other". Justify the statement with suitable example.

The statement

Thevenin's theorem replaces a network by a voltage source V_th in series with R_th. Norton's theorem replaces the same network by a current source I_N in parallel with R_N. The two are duals: voltage and current, series and parallel are interchanged, and one can be converted into the other by source transformation.

 Thevenin                      Norton
 o--[R_th]--o a                o---+------+---o a
   (V_th)                         (I_N)  [R_N]
 o---------o b                 o---+------+---o b

Justification

Quantity	Thevenin	Norton
Source	voltage source V_th	current source I_N
Resistance connection	R_th in series	R_N in parallel
Source value	open-circuit voltage	short-circuit current
Resistance	R_th	R_N = R_th

Relations from the two equivalents (same terminal behaviour):

I_N = V_th/R_th
V_th = I_N R_N
R_N = R_th

Because the two circuits give the same load current for every load, each can replace the other (source transformation).

Example

Network: 12 V source with 4 Ω in series, load R_L = 6 Ω across a-b.

Thevenin: V_th = 12 V, R_th = 4 Ω. I_L = 12/(4 + 6) = 1.2 A.
Norton: I_N = 12/4 = 3 A, R_N = 4 Ω. I_L = 3 × 4/(4 + 6) = 1.2 A.

The load current is the same, and V_th = I_N R_N = 3 × 4 = 12 V confirms the duality.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Thevenin and Norton duality

The statement

Justification

Example

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