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Physics / Chapter 26: Current and Resistance / Section 26-4

Ohm's Law — Question 6

Section 26-4 · Physics pages 4222–4223

Question 6 A symmetric current-voltage curve need not satisfy superposition A device obeys I(V ) = GV + βV 3, with positive G and beta. A proposed test uses two separately applied voltages V1, V2. V I Axes and labels specify the model; the sketch is schematic. (a) Compute I(V1 + V2) −I(V1) −I(V2). (b) For V1 = V2 = V0, find the discrepancy and compare with the polarity-reversal test I(−V ) = −I(V ).
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Question 6 - Solution A symmetric current-voltage curve need not satisfy superposition V I (a) Method and calculation ∆I = β[(V1 + V2)3 −V 3 1 −V 3 2 ]. ∆I = 3βV1V2(V1 + V2). (b) Interpretation and further calculation ∆I(V0, V0) = 6βV 3 0̸ = 0 (V0̸ = 0). I(−V ) = −GV −βV 3 = −I(V ). Odd symmetry is weaker than linearity. Physical check A material can respond symmetrically to reversed polarity yet fail additivity.

Question 6

A symmetric current-voltage curve need not satisfy superposition

A device obeys I(V)=GV+βV3I(V)=GV+\beta V^3, with positive G and beta. A proposed test uses two separately applied voltages V1,V2V_1,V_2.

Diagram shown in the original SVG worksheet.

  1. Compute I(V1+V2)−I(V1)−I(V2)I(V_1+V_2)-I(V_1)-I(V_2).

  2. For V1=V2=V0V_1=V_2=V_0, find the discrepancy and compare with the polarity-reversal test I(−V)=−I(V)I(-V)=-I(V).

Question 6 - Solution

A symmetric current-voltage curve need not satisfy superposition

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔI=β[(V1+V2)3−V13−V23].\Delta I=\beta[(V_1+V_2)^3-V_1^3-V_2^3].ΔI=3βV1V2(V1+V2).\Delta I=3\beta V_1V_2(V_1+V_2).

(b) Interpretation and further calculation

ΔI(V0,V0)=6βV03≠0(V0≠0).\Delta I(V_0,V_0)=6\beta V_0^3\ne0\quad(V_0\ne0).I(−V)=−GV−βV3=−I(V).I(-V)=-GV-\beta V^3=-I(V).Odd symmetry is weaker than linearity.

Physical check

A material can respond symmetrically to reversed polarity yet fail additivity.

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