Calculating Potential and Voltages in Series Capacitors

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SUMMARY

This discussion focuses on calculating the potential and voltages in a series capacitor circuit connected to a 15-V battery. The capacitors involved have capacitances of 4.5 microFarads, 12.0 microFarads, and 32 microFarads. The overall capacitance of the series configuration is determined to be 2.97 microFarads, leading to a charge of 4.45 x 10^-5 C. Consequently, the voltage across the 32 microFarad capacitor is calculated to be 1.39 V, with an alternative method suggested using voltage division by treating the capacitors as impedances.

PREREQUISITES
  • Understanding of series capacitor configurations
  • Knowledge of capacitance calculations
  • Familiarity with the concept of charge conservation in capacitors
  • Basic principles of voltage division in electrical circuits
NEXT STEPS
  • Study the formula for calculating total capacitance in series configurations
  • Learn about the conservation of charge in capacitor circuits
  • Explore voltage division techniques in AC and DC circuits
  • Investigate the use of complex impedance in capacitor analysis
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Electrical engineering students, circuit designers, and anyone involved in analyzing or designing capacitor circuits will benefit from this discussion.

triplezero24
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1. For http://www.pen.eiu.edu/~cuemc4/circuit.bmp picture.. Calculate the potential at point D.

2. I 15-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.5microFarads, 12.0microFarads, and 32microFarads. Find the voltage across the 32microFarad capacitor.

I am so lost. :frown: :confused:
 
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i can't see the circuit
 
1. Can't load the circuit diagram.

2. The absolute value of the charge collected on each capacitor plate will be equal (conservation law of charge), suppose it is Q. The overall capacitance of the three capacitors in series is: (4.5^-1 + 12^-1 + 32^-1)^-1 uF = 2.97 uF (uF = microfarad)

Q = CV
= 2.97 * 10^-6 * 15 C
= 4.45 * 10^-5 C

So the voltage across the 32 uF capacitor is:
V = Q/C
= (4.45 * 10^-5) / (32 * 10^-6) V
= 1.39 V

NB: You could have also used voltage division by treating the capacitances as impedances with values -j/(2pi*fC) ohms, since the 15V DC source can be thought of as a sinusoidal voltage source with infinitesimal frequency.
 
Last edited:
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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