# Calculating net change of capacitor charge in a circuit

• Engineering

## 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 ## 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" ...

Last edited:

## Answers and Replies

gneill
Mentor
Is E3 also an unknown quantity?

I updated the question. E3 is known, not E1. I apologize.

gneill
Mentor
So to confirm, unknown component values are:
E1
R4
R5
R7
R8
C

Yes, that is correct.

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.