Tricky circuit calculating the total capacitance

In summary, to calculate the equivalent capacitance between points a and b in the given figure, we can use the equations for series and parallel combinations. However, since this system is not a simple series or parallel combination, we can use symmetry to simplify the problem. By taking out the 8 uF capacitor and applying a voltage to points a and b, we can determine the voltage between the two C1's and the voltage between the two C2's. This information can then be used to calculate the equivalent capacitance between points a and b.
  • #1
need_aca_help
32
0

Homework Statement


Calculate the equivalent capacitance between points a and b in the figure below (C1 = 4.10 μC and C2 = 1.60 μC). Notice that this system is not a simple series or parallel combination.

cw27A8Q.gif


Homework Equations


Series:
CT = [1/c1 + 1/c2...]-1

Parallel:
CT = C1 + C2...

The Attempt at a Solution


I don't know where to tackle this problem...
 
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  • #2
need_aca_help said:

Homework Statement


Calculate the equivalent capacitance between points a and b in the figure below (C1 = 4.10 μC and C2 = 1.60 μC). Notice that this system is not a simple series or parallel combination.

cw27A8Q.gif


Homework Equations


Series:
CT = [1/c1 + 1/c2...]-1

Parallel:
CT = C1 + C2...

The Attempt at a Solution


I don't know where to tackle this problem...
As the problem states, this is not a combination of series and parallel pieces..

Use symmetry.
 
  • #3
SammyS said:
As the problem states, this is not a combination of series and parallel pieces..

Use symmetry.

Symmetry...? Not sure how to do that...
 
  • #4
OK, hint: take out the 8 uF capacitor, apply any voltage to a and b, what is the voltage between the two C1's and what is the voltage between the two C2's?
 
  • #5

I would approach this problem by first analyzing the circuit and identifying any potential simplifications or patterns. From the given information, I can see that the two capacitors are not in a simple series or parallel configuration. However, I notice that the two capacitors are connected in a way that forms a loop. This suggests that the circuit may be using the concept of a "loop rule", where the sum of the voltage drops around a closed loop is equal to zero.

Using this information, I would then apply Kirchhoff's loop rule to the circuit, writing out the equations for the voltage drops across each capacitor. From there, I would use the equations to solve for the unknown voltage drops and then use the formula for capacitance (C = Q/V) to calculate the equivalent capacitance between points a and b.

Additionally, I would double-check my calculations and make sure they are consistent with the given values for C1 and C2. If my calculated equivalent capacitance is significantly different from the given values, I would re-evaluate my approach and check for any mistakes. As a scientist, it is important to always double-check and verify our results to ensure accuracy and precision in our work.
 

Related to Tricky circuit calculating the total capacitance

1. How do I calculate the total capacitance of a tricky circuit?

To calculate the total capacitance of a tricky circuit, you need to first identify all the capacitors in the circuit. Then, use the formula Ceq = C1 + C2 + C3 + ... to add up the individual capacitances. Make sure to keep in mind the parallel and series connections in the circuit and use the appropriate formula for each. Finally, simplify the equation to get the final value for the total capacitance.

2. What is the difference between capacitors in series and parallel?

Capacitors in series have a combined capacitance that is less than that of each individual capacitor. This is because the total capacitance is limited by the smallest capacitor in the series. On the other hand, capacitors in parallel have a combined capacitance that is greater than that of each individual capacitor. This is because the total capacitance is the sum of all the individual capacitances.

3. Can I use the total capacitance to calculate the charge stored in a circuit?

Yes, you can use the total capacitance to calculate the charge stored in a circuit. The formula for charge is Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. By rearranging this formula, you can calculate the charge using the total capacitance and the voltage of the circuit.

4. How does the placement of capacitors affect the total capacitance in a circuit?

The placement of capacitors in a circuit can affect the total capacitance in different ways. In series, the total capacitance decreases as more capacitors are added. In parallel, the total capacitance increases as more capacitors are added. The placement of capacitors also affects the overall impedance of the circuit, which can impact the flow of current.

5. Can I use the total capacitance to calculate the time constant of a circuit?

Yes, you can use the total capacitance to calculate the time constant of a circuit. The time constant is a measure of how quickly the voltage across a capacitor changes. It is calculated by multiplying the total capacitance by the resistance in the circuit. This value can be used to determine the rate at which the capacitor charges and discharges in the circuit.

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