- #1
diredragon
- 321
- 15
Homework Statement
This is the second problem from our test preparation cycle and i partially solved and understood it.
The circuit in the picture is in the state of switch being open (STATE 1: SWITCH OPEN). Calculate ##E1## so that the change in the electrical energy of the capacitor is ##ΔW=4uJ##.
Homework Equations
3. The Attempt at a Solution [/B]
##ΔW=1/2ΔU^2_{1/2}C=4uJ##
##ΔU^2_{1/2}=\pm 2V## [there are two values and we can't determine which is correct so we work both]
##ΔU=ΔU_{ab}## [that voltage is the same for points A and B]
Starting to simplify the circuit and added current generator to represent the change in that branch of the circuit:
##Δ=R1+R2+R3## [starting to transfer a triangle into a star]
##RA=\frac{R1R2}{Δ}##;##RB=\frac{R1R3}{Δ}##;##RC=\frac{R2R3}{Δ}## [finished with that]
##ΔU_{ab}=a*ΔIGK## [figured its a superposition principle where we find Uab if only IGK works]
##ΔIGK=\pm 2/55 mA## [immediate jump to the result, didnt quite get that, how did they get that? I am going to try to derive it myself] [dont get #1]
TRY:
##I_{branch}=ΔIGK*\frac{RB+R5}{RA+RB+R4+R5}## [tried the current divider to get the current in left part]
##U_{ab}=I_{branch}*(R4+RA)+ΔIGK*RC## [didnt check the results but i will, does this seem good though?]
Continuation of the solution:
##ET=E\mp IR=\mp 2V## [so the theorem of compensation, putting the voltage source instead of the whole branch]
##I=ΔIGK## [the change that happens is that the current flows through the branch]
##ET'=E1/8## [now this all of a sudden, how did they get to this?] [dont get #2]
##RT'=55## [the equivalent thevenin resistance i get this but the upper part i can't get still]
##ET'-ET+RT'I=0## [why +RT'I and not -RT'I?] [dont get #3]
##ET'=0 \Rightarrow E1=0##
##ET'0=4 \Rightarrow E1=32##