Calculating net change of capacitor charge in a circuit

In summary, the problem involves calculating the net change of capacitor charge in a circuit with given parameters and known current flow when the switch is closed. Using various circuit theorems such as Ohm's law, current divider, Thevenin and Norton theorems, and Millman's theorem, we can determine the voltage and charge at the ends of the capacitor in both the closed and open switch states. By finding the proportionality constant between voltage and charge, we can calculate the net change of capacitor charge to be -0.6 µC.
  • #1
irrationally
11
0

Homework Statement


This problem apparently has a simple solution if you are good with circuit theorems but no matter what i try or how i combine them, it's just impossible. Given a circuit on the picture, and circuit parameters E3=12V,R1=1kΩ,R2=30Ω,R3=150Ω,R6=200Ω, if it is known that current I1 = 60 mA when the switch (P) is closed, calculate the net change of capacitor charge between switch states ( switch is first closed, and then opened - there is also a time delay between switch changes so we are only interested in stationary states).
EDIT: Capacitor charge for when switch is closed is Q=-2 µC
EDIT: Thanks gneill, I made a mistake. E3 is known, not E1

5r31L.png


Homework Equations


Ohm's law - U = R*I
Current devider
Thevenein and Norton theorem
Superposition theorem
Node potential method
Mesh current method ( i doub't this is useful in this example )[/B]

The Attempt at a Solution


I did try but i just can't get anything useful. I don't even have capacitor capacitance so i can't calculate voltage (Uab). Millman's theorem came to mind ( for when the switch is open, we can calculate voltage U1 in at the end of edge with E3, and then compensate that edge with an ideal voltage generator with EMS equal to U1. From here, since there is no current flow trough the capacitor, we have a single generator connected to a resistor network but we don't know 4 out of 5 resistors' resistances) but it seems like a dead end. As for the situation with switch closed, that's even worse. No matter what i do, I just can;t "reach the capacitor" ...
 
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  • #2
Is E3 also an unknown quantity?
 
  • #3
I updated the question. E3 is known, not E1. I apologize.
 
  • #4
So to confirm, unknown component values are:
E1
R4
R5
R7
R8
C
 
  • #5
Yes, that is correct.
 
  • #6
I solved the task.
First i considered the case with the closed switch
First edge ( all orientation is from the left) can be replaced with an ideal current generator with current I= I1. Now, edge one and two (which form a Norton generator) can be converted into a Thevenin generator with ems Et =I1R2, and resistance R2. Now, by using Millman's theorem, we can find the voltage Um at the ends of edge with E3. We can compensate this edge (and everything left from it ) with a compensatory ideal voltage generator with ems Ec = Um. Theorem of proportionality says that voltage Uc ( voltage at the ends of capacitor) is proportional to Ec i.e. Uc=a*Ec. We don't have this voltage but since Q = C*U, and capacitance C is constant, we can conclude that charge is also proportional to Ec i.e. Qc = b*Ec. From here we can calculate the proportionality constant b.

Now the case with switch closed ( all values that have changed are marked with ' , e.g. U' :
There is no current flow trough edge one so it can be compensated with a current generator with current 0. We do not need to look at this edge any more. Now , again, we use Millman's theorem for to find the Um' andn eventually Ec'. This is all we need and now we can simply calculate Qc' = b * Ec' . This yeald a result of Qc' - Qc = -0.6 µC.
 

1. How do you calculate the net change of capacitor charge in a circuit?

The net change of capacitor charge in a circuit can be calculated by subtracting the initial charge on the capacitor from the final charge on the capacitor. This value represents the change in charge that has occurred in the capacitor.

2. What is the formula for calculating the net change of capacitor charge?

The formula for calculating the net change of capacitor charge is Q = C x (Vf - Vi), where Q is the net change in charge, C is the capacitance of the capacitor, Vf is the final voltage on the capacitor, and Vi is the initial voltage on the capacitor.

3. How does the capacitance affect the net change of charge in a capacitor?

The capacitance of a capacitor is directly proportional to the net change of charge in a circuit. This means that as the capacitance increases, the net change of charge also increases. This is because a higher capacitance allows the capacitor to store more charge.

4. Can the net change of capacitor charge be negative?

Yes, the net change of capacitor charge can be negative. This occurs when the final voltage on the capacitor is lower than the initial voltage, resulting in a decrease in charge stored in the capacitor. In this case, the net change of charge will have a negative value.

5. What are some real-world applications of calculating the net change of capacitor charge in a circuit?

The net change of capacitor charge is an important concept in many electronic devices. It is used in the design and operation of circuits such as power supplies, audio amplifiers, and camera flashes. It is also essential in calculating the energy stored in capacitors, which is crucial for the efficient use of energy in electronic systems.

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