Find current for circuit with capacitor, resistor, capacitor in series

Click For Summary
SUMMARY

The discussion focuses on solving the current function of time for a circuit comprising two capacitors (1uF each) and a resistor (100k-Ohm) in series. The user applied Kirchhoff's Voltage Law (KVL) to derive the equation but encountered an issue when simplifying the equation due to the equal capacitances leading to an incorrect equivalent resistance (RC-eq). The user suggests that changing the sign of one capacitor or adjusting the combined capacitance to 0.5uF may resolve the issue, as the configuration is passive.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of series circuits with capacitors and resistors
  • Familiarity with differential equations in electrical engineering
  • Concept of equivalent capacitance in passive configurations
NEXT STEPS
  • Study the application of Kirchhoff's Voltage Law in series circuits
  • Learn about the behavior of capacitors in series and their equivalent capacitance
  • Explore differential equations related to RC circuits
  • Investigate passive circuit configurations and their implications on circuit analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing series circuits with capacitors and resistors.

Cyrus
Messages
3,246
Reaction score
17
I have to solve this problem for a circuit. Its a capacitor, a resistor and another capacitor all in series.

Both capacitances are the same, 1uF, and the resistor is 100k-Ohm.

I have to find the current function of time.

What I did was use KVL to get:


[tex]- \frac{1}{c_1} \int i(t)dt + v_1 (t_0) + \frac{1}{c_2} \int i(t)dt + v_2 (t_0) + R i(t) = 0[/tex]

Then I took the derivative to get:

[tex]- \frac{1}{c_1} i(t) + R \frac {di}{dt} + \frac{1}{c_2}i(t) =0[/tex]

Which simplifies to:

[tex]R( \frac{1}{c_2} - \frac {1}{c_1} )^{-1} \frac {di}{dt} +i(t) = 0[/tex]

But the two capacitances have the same value, which means that RC-eq is zero. That's wrong...hmmmmmmmmm
 
Last edited:
Physics news on Phys.org
WTF. Why not change the signs of one of the capacitances. They should have the same sign. Or just let the combined capacitance equal to .5uF.
 
Because based on the diagram, they are in the passive configuration which is why it is negative.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
Replies
20
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K