Ohm's Law — Question 3

Section 26-4 · Physics pages 4216–4217

Question 3 Two operating points identify the onset of nonohmic response A device follows I = GV + βV 3. At V=1.00 V its current is 3.00 mA; at V=2.00 V it is 12.0 mA. Temperature is held constant. V I Axes and labels specify the model; the sketch is schematic. (a) Infer G and beta. (b) Find the nonzero voltage magnitude at which the cubic current equals the linear current, and the differential conductance there.
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Question 3 - Solution Two operating points identify the onset of nonohmic response V I (a) Method and calculation G + β(1 V)2 = 3.00 mS, G + β(2 V)2 = 6.00 mS. G = 2.00 mS, β = 1.00 mA/V3. (b) Interpretation and further calculation βV 3 ∗= GV∗⇒|V∗| = √ G/β = √ 2 V. gd = G + 3βV 2 ∗= 4G = 8.00 mS. Physical check The extra current cannot be explained by a single constant resistance at both operating points.

Question 3

Two operating points identify the onset of nonohmic response

A device follows I=GV+βV3I=GV+\beta V^3. At V=1.00 V its current is 3.00 mA; at V=2.00 V it is 12.0 mA. Temperature is held constant.

Diagram shown in the original SVG worksheet.

  1. Infer G and beta.

  2. Find the nonzero voltage magnitude at which the cubic current equals the linear current, and the differential conductance there.

Question 3 - Solution

Two operating points identify the onset of nonohmic response

Diagram shown in the original SVG worksheet.

(a) Method and calculation

G+β(1V)2=3.00mS,G+β(2V)2=6.00mS.G+\beta(1\,\mathrm V)^2=3.00\,\mathrm{mS},\quad G+\beta(2\,\mathrm V)^2=6.00\,\mathrm{mS}.G=2.00mS,β=1.00mA/V3.G=2.00\,\mathrm{mS},\quad\beta=1.00\,\mathrm{mA/V^3}.

(b) Interpretation and further calculation

βV*3=GV*⇒|V*|=G/β=2V.\beta V_*^3=GV_*\Rightarrow |V_*|=\sqrt{G/\beta}=\sqrt2\,\mathrm V.gd=G+3βV*2=4G=8.00mS.g_d=G+3\beta V_*^2=4G=8.00\,\mathrm{mS}.

Physical check

The extra current cannot be explained by a single constant resistance at both operating points.