Finding Vc for two circuits, Electric Circuit Analysis question

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Discussion Overview

The discussion revolves around finding the voltage across a capacitor (Vc) in two different electric circuit scenarios, specifically focusing on circuit analysis techniques. Participants explore the implications of circuit configurations on the behavior of the capacitor in both cases.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the capacitor is not in parallel with the current source in case (a), suggesting a need for recalculation of Vc in both cases.
  • In case (b), one participant calculates the Thevenin resistance (Rth) as 15.04 ohms and finds Vc to be 67.68 nV.
  • For case (a), another participant mentions that the capacitor acts as a short circuit in a DC circuit, leading to an equivalent resistance (Req) of 7.37 ohms and a voltage of 33.16 nV, but expresses uncertainty about the next steps.
  • Some participants agree that the voltage across the capacitor in case (a) should equal the voltage across the 7 ohm resistor and inquire about the current flowing through it.
  • One participant describes a method for calculating equivalent resistance in both cases, emphasizing that the capacitor is treated as open-circuited during these calculations.

Areas of Agreement / Disagreement

Participants generally agree on the calculations for case (b) but exhibit disagreement and uncertainty regarding the analysis of case (a), with multiple interpretations of the circuit configuration and its implications for Vc.

Contextual Notes

Participants have not resolved the assumptions regarding the circuit configurations, particularly in case (a), and there are unresolved mathematical steps related to the current through the resistors.

Xiao Xiao
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Homework Statement
For each circuit shown in Fig. 7.45, calculate the voltage labeled vc.
Relevant Equations
In both circuits I should consider an open circuit in place of the capacitor, and then calculate Req for the remaining resistors (5ohm resistor is ommited in circuit-a) then use v=iR to find the the voltage, and by parallel v=vc (capacitor's voltage) is my answer correct? I'm sorry if this is a silly question.
IMG_20210504_141141.png
 
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I think what you said is correct in case (b), but not in case (a). The capacitor is not in parallel with the current source in case (a). Why don't you try calculating vc in both cases and show us your work?
 
phyzguy said:
I think what you said is correct in case (b), but not in case (a). The capacitor is not in parallel with the current source in case (a). Why don't you try calculating vc in both cases and show us your work?
For (b) I did Rth= 15.04ohm, Vc=v=iR=67.68nV

For (a) capacitor is short circuit in dc circuit, so we ignore the 5ohm so Req=7. 37ohm, and v=33.16nV, if what you're saying is correct then I'm not sure how to proceed after that.
 

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Part (b) looks correct. For part (a), the voltage across the capacitor will be equal to the voltage across the 7 Ohm resistor. How much current is flowing through the 7 Ohm resistor, and what s the voltage across it?
 
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phyzguy said:
Part (b) looks correct. For part (a), the voltage across the capacitor will be equal to the voltage across the 7 Ohm resistor. How much current is flowing through the 7 Ohm resistor, and what s the voltage across it?
Oooh, I see, thanks a lot, I'll try to do that now.
 
In both cases the capacitor is open circuited when the equivalent resistance is calculated as follows. In case (a) you have 17 ohms in parallel with 13 ohms. Collapse that into one resistor and place it in parallel with the open circuited capacitor and just do V=IR. In case (b) you have 18 ohms in parallel with 7 ohms and then in series with 10 ohms. Collapse that into one resistor in parallel with the open circuited capacitor and, once again, do V=IR.
 
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