Potential difference across capacitor in a network

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SUMMARY

The discussion centers on determining the potential difference across a capacitor in a circuit using Kirchhoff's loop rule and the relationship between charge, capacitance, and voltage. Participants clarify that the equations V1 = -q/c1 and V2 = q/c2 can yield different signs based on the reference point for ground potential. The conversation highlights the ambiguity in the question's wording from the All India CBSE board 2013 exam, leading to confusion regarding voltage polarity. Ultimately, multiple valid interpretations of the voltage signs are presented, emphasizing the importance of context in circuit analysis.

PREREQUISITES
  • Understanding of Kirchhoff's loop rule
  • Familiarity with capacitor equations: V = q/c
  • Knowledge of electrical potential and reference points
  • Basic circuit analysis skills
NEXT STEPS
  • Study the implications of reference points in circuit analysis
  • Explore advanced applications of Kirchhoff's laws in complex circuits
  • Learn about different types of capacitors and their behaviors in circuits
  • Investigate common pitfalls in interpreting exam questions related to electrical circuits
USEFUL FOR

Students preparing for electrical engineering exams, educators developing circuit analysis curriculum, and anyone interested in understanding capacitor behavior in electrical networks.

Suyash Singh

Homework Statement


Determine the potential difference across the plates of the capacitor of the network shown in the figure.

6cfd021e59d9eff49396f757df312d4af4ab835d.png

Homework Equations


kirchoff's loop rule
q=vc
3. The attempt at a solutio
applying kirchhoffs law
starting at bottom right point,
q/c2-E1+q/c1+E2=0
=>E2-E1=-q/c1-q/c2
now i am confused what to do further

the answer is
V1=-q/c1 and V2=q/c2
 
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You don't need Kirchoff's law. Look at the second of the two relevant equations that you wrote down. Give q and c, what is v?
 
phyzguy said:
You don't need Kirchoff's law. Look at the second of the two relevant equations that you wrote down. Give q and c, what is v?
v=q/c
but why is v1 negative and v2 positive?
 
Suyash Singh said:
v=q/c
but why is v1 negative and v2 positive?
Negative or positive with respect to what?
Looks like they are assuming the ground to be at the negative terminal of E2.
 
+1

I think it is a badly written question. Is that the exact wording?
 
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CWatters said:
+1

I think it is a badly written question. Is that the exact wording?
yes it was asked in All India CBSE board 2013.
 
cnh1995 said:
Negative or positive with respect to what?
Looks like they are assuming the ground to be at the negative terminal of E2.
the ground is taken at zero potential in our syllabus.
 
Even if the 0V node was specified the polarity of the voltages isn't defined. Consider this circuit...

Polarity.png


If E1 = 10V then V3 = 10V and V2 = -10V. See how the polarity depends on the direction of the arrow I drew on the circuit.

So I would argue that in your circuit all of the following are valid answers...

V1=+q/c1 and V2=+q/c2
V1=-q/c1 and V2=+q/c2
V1=+q/c1 and V2=-q/c2
V1=-q/c1, and V2=-q/c2
 
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