Calculating Potential Difference on A Capacitor, RC Circuits

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SUMMARY

The potential difference across a capacitor in an RC circuit can be determined using Kirchhoff's Voltage Law. In the discussed circuit with a battery voltage of 31 V and a capacitor of 9 µF, the fully charged capacitor will not have a potential difference equal to the battery voltage due to the presence of resistors in the circuit. The current flowing through the resistors must be calculated to find the voltage across the capacitor, which is influenced by the series resistors of 2 ohms and 3 ohms. The voltage drop across the 3-ohm resistor can be used to ascertain the voltage on the capacitor, as there is no current through the 1-ohm resistor.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law
  • Basic knowledge of RC circuit behavior
  • Familiarity with Ohm's Law (V=IR)
  • Concept of capacitor charging and discharging
NEXT STEPS
  • Calculate the current in an RC circuit using Ohm's Law
  • Explore the behavior of capacitors in series and parallel configurations
  • Learn about transient analysis in RC circuits
  • Study the effects of different resistor values on capacitor charging time
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in understanding the behavior of capacitors in RC circuits.

cwatki14
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The circuit of Figure P.72 has been turned on for a long time, so that you can ignore the transient behavior of the capacitor. Voltage of Battery V = 31 V
p19-72alt.gif


(e) What is the potential difference across the capacitor C (C = 9 µF)?

I thought about using Kirchoff Voltage Law, but I also though that when a capacitor is fully charged the potential difference should be equal to that on the battery, but the computer said that was wrong. It also is not zero. Any ideas?
 
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Initially the uncharged capacitor will have very voltage, so a lot of current will flow that way. But as it becomes charged its voltage rises until no current at all flows into it. The resistance of that part of the circuit is effectively infinite, so you just ignore it. The circuit you have left is the battery driving current into the 2 and 3 ohm resistors in series. Calculate that current and the voltage across the 3 ohm resistor. Since there is no current through the 1 ohm resistor, it will have no voltage drop V=IR across it and so you know the voltage on the capacitor.
 

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