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Other Single-Loop Circuits — Question 6

Section 27-4 · Physics pages 4382–4383

Question 6 A threshold element selects which algebraic loop solution is admissible A device conducts only from left to right. In that direction its voltage drop is Vt + rdI for I > 0; at I = 0 it can support any applied voltage below Vt, including negative values. It is in series with resistor R and signed source emf E. + − E R Vt + rdI I Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find I as a piecewise function of E. (b) For Vt = 2.00 V, rd = 10.0 Ωand R = 90.0 Ω, find I at E=12.0 V and E=1.00 V.
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Question 6 - Solution A threshold element selects which algebraic loop solution is admissible + − E R Vt + rdI I (a) Method and calculation E −IR −(Vt + rdI) = 0 ⇒I = (E −Vt)/(R + rd) if E > Vt. I(E) = { 0, E ≤Vt, (E −Vt)/(R + rd), E > Vt. (b) Interpretation and further calculation I(12.0 V) = (12 −2)/100 = .100 A. I(1.00 V) = 0; (1 −2)/100 < 0 violates the conducting-branch condition. Physical check The loop equation must be solved together with the device operating regime; a negative branch current is not admissible for this idealized element.

Question 6

A threshold element selects which algebraic loop solution is admissible

A device conducts only from left to right. In that direction its voltage drop is Vt+rdIV_t+r_dI for I>0I>0; at I=0I=0 it can support any applied voltage below Vt, including negative values. It is in series with resistor R and signed source emf E.

Diagram shown in the original SVG worksheet.

  1. Find I as a piecewise function of E.

  2. For Vt=2.00V_t=2.00 V, rd=10.0Ωr_d=10.0\,\Omega and R=90.0ΩR=90.0\,\Omega, find I at E=12.0 V and E=1.00 V.

Question 6 - Solution

A threshold element selects which algebraic loop solution is admissible

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E−IR−(Vt+rdI)=0⇒I=(E−Vt)/(R+rd)if E>Vt.E-IR-(V_t+r_dI)=0\Rightarrow I=(E-V_t)/(R+r_d)\quad\text{if }E>V_t.I(E)={0,E≤Vt,(E−Vt)/(R+rd),E>Vt.I(E)=\begin{cases}0,&E\le V_t,\\(E-V_t)/(R+r_d),&E>V_t.\end{cases}

(b) Interpretation and further calculation

I(12.0V)=(12−2)/100=.100A.I(12.0\,\mathrm V)=(12-2)/100=.100\,\mathrm A.I(1.00V)=0;(1−2)/100<0 violates the conducting-branch condition.I(1.00\,\mathrm V)=0;\quad (1-2)/100<0\text{ violates the conducting-branch condition}.

Physical check

The loop equation must be solved together with the device operating regime; a negative branch current is not admissible for this idealized element.

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