RC Circuits — Question 3

Section 27-8 · Physics pages 4456–4457

Question 3 Three equally spaced samples reveal both the time constant and the unknown final voltage A first-order RC voltage follows v(t) = V∞+ Ae−t/τ. Ideal samples give v(0)=2.00 V, v(.100 s)=6.00 V and v(.200 s)=8.00 V. The driving resistance is 10.0 kilohm. t (s) vC (V) 2 6 8 0 .100 .200 Schematic; labels, not drawing lengths, specify the data. (a) Eliminate the unknown offset to find tau and V infinity. (b) Find C and write the complete voltage function; verify all three samples.
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Question 3 - Solution Three equally spaced samples reveal both the time constant and the unknown final voltage t (s) vC (V) 2 6 8 0 .100 .200 (a) Method and calculation a = e−∆t/τ = v2 −v1 v1 −v0 = 2 4 = 1 2, τ = .100 ln 2 = .14427 s. V∞= v1 −av0 1 −a = 10.0 V. (b) Interpretation and further calculation C = τ/R = 14.427 µF, v(t) = 10 −8e−t/τ V. v(n∆t) = 10 −8(1/2)n : v0 = 2, v1 = 6, v2 = 8 V. Physical check Successive voltage differences remove the unknown final value, provided one exponential describes the response.

Question 3

Three equally spaced samples reveal both the time constant and the unknown final voltage

A first-order RC voltage follows v(t)=V∞+Ae−t/τv(t)=V_\infty+A e^{-t/\tau}. Ideal samples give v(0)=2.00 V, v(.100 s)=6.00 V and v(.200 s)=8.00 V. The driving resistance is 10.0 kilohm.

Diagram shown in the original SVG worksheet.

  1. Eliminate the unknown offset to find tau and V infinity.

  2. Find C and write the complete voltage function; verify all three samples.

Question 3 - Solution

Three equally spaced samples reveal both the time constant and the unknown final voltage

Diagram shown in the original SVG worksheet.

(a) Method and calculation

a=e−Δt/τ=v2−v1v1−v0=24=12,τ=.100ln⁡2=.14427s.a=e^{-\Delta t/\tau}=\frac{v_2-v_1}{v_1-v_0}=\frac24=\frac12,\quad\tau=\frac{.100}{\ln2}=.14427\,\mathrm s.V∞=v1−av01−a=10.0V.V_\infty=\frac{v_1-av_0}{1-a}=10.0\,\mathrm V.

(b) Interpretation and further calculation

C=τ/R=14.427μF,v(t)=10−8e−t/τV.C=\tau/R=14.427\,\mathrm{\mu F},\quad v(t)=10-8e^{-t/\tau}\ \mathrm V.v(nΔt)=10−8(1/2)n:v0=2,v1=6,v2=8V.v(n\Delta t)=10-8(1/2)^n:\quad v_0=2,\ v_1=6,\ v_2=8\ \mathrm V.

Physical check

Successive voltage differences remove the unknown final value, provided one exponential describes the response.