Capacitors inserted into a circuit with opposite polarities

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SUMMARY

The discussion centers on the behavior of two capacitors, C1 = 5 μF and C2 = 2 μF, connected in parallel with opposite polarities. When the switches S1 and S2 are closed, the potential difference between points A and B is not zero, contrary to initial assumptions. This is due to the unequal charges on the connected plates, which results in a net voltage across the circuit. The correct approach involves calculating the total charge and resulting voltage from the capacitors.

PREREQUISITES
  • Understanding of capacitor behavior in circuits
  • Knowledge of parallel circuit configurations
  • Familiarity with the formula C = Q / ΔV
  • Basic principles of electric charge and potential difference
NEXT STEPS
  • Calculate the total charge for capacitors in parallel
  • Explore the effects of capacitor polarity on circuit behavior
  • Learn about voltage division in capacitor circuits
  • Study the implications of unequal capacitance in parallel configurations
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone interested in understanding capacitor behavior in parallel circuits.

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Homework Statement



Two capacitors C1 = 5 μF and C2 = 2 μF are initially charged to a potential difference of V = 100 V. They are inserted into the circuit with opposite polarities as shown. Switches S1 and S2 are now closed. What is the magnitude of the potential difference between points A and B? (Answer in Volts.)

[PLAIN]http://img833.imageshack.us/img833/557/phyr.jpg

Homework Equations



C = Q / deltaV

The Attempt at a Solution



Since both capacitors have same potential difference since they are connected in parallel. I think the answer should be zero since we would have two opposite-sign voltages.
 
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No. The voltage will not be zero as the charges are not equal and opposite on the connected plates.

ehild
 
No. The voltage will not be zero as the charges are not equal and opposite on the connected plates.

ehild
 

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