Capacitors in Parallel and in Series — Question 4

Section 25-3 · Physics pages 4058–4059

Question 4 Series and parallel measurements determine an unknown pair up to interchange Two positive capacitances have parallel equivalent 10.0 nF and series equivalent 2.40 nF. A B C1 C2 M Schematic; labels, not drawing lengths, specify the data. (a) Find the individual capacitances. (b) Derive the condition on general measured values Cp and Cs for two real positive capacitances to exist.
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Question 4 - Solution Series and parallel measurements determine an unknown pair up to interchange A B C1 C2 M (a) Method and calculation C1 + C2 = 10.0, C1C2 = CsCp = 24.0 in nF units. C2 −10C + 24 = 0 ⇒{C1, C2} = {4.00, 6.00} nF. (b) Interpretation and further calculation C1,2 = Cp ± √ C2p −4CpCs 2 . Cp > 0, 0 < Cs ≤Cp/4. Physical check Equality corresponds to identical capacitors; the two measurements cannot identify which physical device has which value.

Question 4

Series and parallel measurements determine an unknown pair up to interchange

Two positive capacitances have parallel equivalent 10.0 nF and series equivalent 2.40 nF.

Diagram shown in the original SVG worksheet.

  1. Find the individual capacitances.

  2. Derive the condition on general measured values Cp and Cs for two real positive capacitances to exist.

Question 4 - Solution

Series and parallel measurements determine an unknown pair up to interchange

Diagram shown in the original SVG worksheet.

(a) Method and calculation

C1+C2=10.0,C1C2=CsCp=24.0in nF units.C_1+C_2=10.0,\quad C_1C_2=C_sC_p=24.0\quad\text{in nF units}.C2−10C+24=0⇒{C1,C2}={4.00,6.00}nF.C^2-10C+24=0\Rightarrow\{C_1,C_2\}=\{4.00,6.00\}\,\mathrm{nF}.

(b) Interpretation and further calculation

C1,2=Cp±Cp2−4CpCs2.C_{1,2}=\frac{C_p\pm\sqrt{C_p^2-4C_pC_s}}2.Cp>0,0<Cs≤Cp/4.C_p>0,\quad0<C_s\le C_p/4.

Physical check

Equality corresponds to identical capacitors; the two measurements cannot identify which physical device has which value.