Voltage generators in open edges and Thevenin's generator

In summary, the problem at hand is to determine the current of an ideal current source in a circuit with known parameters and a capacitor. The use of Thevenin's theorem is optional. The open branches in the circuit have no effect and can be ignored. The strategy for finding the current involves finding the steady-state voltages across the capacitor and relating them using the change in capacitor voltage. The capacitor can be treated as an open circuit during steady state analysis.
  • #1
irrationally
11
0

Homework Statement


2015_02_03_1905.png

I hope you can see the picture clearly. Someone drew lines over it,so please try to ignore that.
The problem here is to determine the current of ideal current source Ig, if following parameters are known:
-E1=50V, E3=30V
-R1=200Ω, R2=500Ω, R3=3kΩ, R4= 10 kΩ
-C=0.7 μF
-Charge flow trough the edge with capacitor after closing the switch Π is Q12 = 0.213 mC.

Homework Equations


-Thevenin and Norton theorem.
-Ohm's law
-Millman's theorem

The Attempt at a Solution


My attempt is to tranform the current generator to a voltage generator and then find the voltage U12. Then i would find Thevenin's generator and equivalent resistance and get a simple circuit. What to do after closing the switch. The biggest problem here is not exactly related to this problem - How does a voltage generator in an open edge affect the rest of the circut? Do we need to take it into calculations? How to set Kirchoff's second law equations? Are resistors in open edges included into calculating equivalent resistance for Thevenin's generator? These problems arise because my base knowledge of circuits is weak. :L:L:L
 
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  • #2
Does the problem statement specifically require you to use Thevenin's Theorem here? I can where nodal analysis might be advantageous.

Regarding open branches (E3 and R3 in your diagram when the switch is open), they have no effect on the circuit. Current cannot flow without a closed path. You may ignore them unless you need the potential at the open switch.
 
  • #3
No, it does not. I'm allowed to solve it any way i want, but i mentioned Thevenin's theorem since i have confusions related to it and i tried using it. Any solution is welcome ;)
 
  • #4
irrationally said:
No, it does not. I'm allowed to solve it any way i want, but i mentioned Thevenin's theorem since i have confusions related to it and i tried using it. Any solution is welcome ;)
I see. Well, you could use Thevenin, but it may be more effort than necessary. Your choice.

What's your overall strategy for finding the current value?
 
  • #5
Well first i find the Thevenin's EMF ( in function of Ig). Then i close the switch and now the voltage of the edge with R3 is equal to the voltage of the capacitor when the switch is open + Δq*C. There is no current flow trough the capacitor so we are left with a simple circuit with elements : E1, R5, Rt, Et(in function of Ig). From here , I hoped to somehow get the Et, but i am kinda stuck.
 
  • #6
irrationally said:
Well first i find the Thevenin's EMF ( in function of Ig). Then i close the switch and now the voltage of the edge with R3 is equal to the voltage of the capacitor when the switch is open + Δq*C. There is no current flow trough the capacitor so we are left with a simple circuit with elements : E1, R5, Rt, Et(in function of Ig). From here , I hoped to somehow get the Et, but i am kinda stuck.
Yeah, you probably want to start by finding expressions for the two steady-state voltages that will appear across the capacitor (both will involve Ig and some constants). Then you can use your change in capacitor voltage Δq*C to relate them. Essentailly that means find the potentials at nodes 1 and 2 for both cases.
 
  • #7
Okay, i can find potentials at nodes 1 and 2 in the first case easily. But what do i do in the second case? How do I treat capacitor when using node potential method for example? My idea is that that branch doesn't contribute to overall current entering the node 0 or 1 , so it can be omitted in calculations. Is this correct?
 
  • #8
irrationally said:
Okay, i can find potentials at nodes 1 and 2 in the first case easily. But what do i do in the second case? How do I treat capacitor when using node potential method for example? My idea is that that branch doesn't contribute to overall current entering the node 0 or 1 , so it can be omitted in calculations. Is this correct?
Right. At steady state no current flows through the capacitor: it looks like an open circuit.
 
  • #9
Oh well this then becomes a very simple problem. Thanks so much.
 

1. What is a voltage generator in open edges?

A voltage generator in open edges is a type of electrical circuit that is used to supply a voltage or potential difference between two open or unconnected points. It is typically represented by a symbol with a plus and minus sign, indicating the direction of the voltage flow.

2. How does a voltage generator in open edges work?

A voltage generator in open edges works by converting a source of energy, such as chemical, mechanical, or solar, into electrical energy. This energy is then used to create a potential difference between two open points, which can be used to power electrical devices.

3. What is Thevenin's generator?

Thevenin's generator is a type of voltage generator that simplifies complex circuits into an equivalent circuit with a single voltage source and a single resistor. It is named after French scientist Leon Charles Thevenin, who developed the concept in the late 19th century.

4. How is Thevenin's generator different from a regular voltage generator?

The main difference between Thevenin's generator and a regular voltage generator is that Thevenin's generator simplifies a complex circuit into a single voltage source and resistor, while a regular voltage generator may have multiple sources and components. Thevenin's generator is also useful for analyzing and understanding the behavior of a circuit.

5. What are the advantages of using Thevenin's generator?

Thevenin's generator has several advantages, including simplifying complex circuits, making it easier to analyze and understand, and providing a way to calculate the voltage and current in a circuit without having to use complex equations. It also allows for easier troubleshooting and designing of circuits.

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