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

In summary, the problem involves solving for the current function of time in a circuit with two capacitors and one resistor in series. The capacitances are both 1uF and the resistor is 100k-Ohm. The solution involves using Kirchhoff's Voltage Law and taking the derivative to simplify the equation. However, the two capacitances have the same value, causing the RC-equivalent to be zero. To correct this, one of the capacitances needs to have its sign changed or the combined capacitance should be adjusted to 0.5uF. The negative value in the equation is due to the passive configuration shown in the diagram.
  • #1
Cyrus
3,238
16
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
 
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  • #2
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.
 
  • #3
Because based on the diagram, they are in the passive configuration which is why it is negative.
 

1. How do I calculate the total current in a circuit with a capacitor, resistor, and capacitor in series?

The total current in this type of circuit can be calculated using Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the total resistance (R). In this case, the total resistance is the sum of the individual resistances of the resistor and the capacitors in series. Therefore, the equation would be I = V / (Rresistor + Rcapacitor1 + Rcapacitor2).

2. What is the effect of the capacitors in series on the current in the circuit?

Capacitors in series have a cumulative effect on the total resistance of the circuit, which in turn affects the current. The capacitors act as insulators, storing charge and reducing the flow of current. This results in a decrease in the total current in the circuit compared to a circuit with only a resistor.

3. How do I find the individual currents through each component in the circuit?

To find the individual currents through each component, you can use the same equation as in question 1, but with the individual resistance values for each component. For example, to find the current through the first capacitor, the equation would be Icapacitor1 = V / Rcapacitor1.

4. Can the total current ever be greater than the current through the resistor?

No, the total current in the circuit will always be equal to or less than the current through the resistor. This is because the capacitors in series act as insulators, limiting the flow of current in the circuit.

5. How does the frequency of the alternating current (AC) affect the current in this type of circuit?

The frequency of the AC does not have a significant effect on the current in this type of circuit. However, it can affect the capacitive reactance of the capacitors, which can then affect the total resistance and current in the circuit. Higher frequencies result in lower capacitive reactance, which in turn increases the total resistance and decreases the current.

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