How do we find the equivalent voltage in a parallel circuit?

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To find the equivalent voltage in a parallel circuit, it's essential to apply Kirchhoff's rules rather than simply adding the voltages together. Assigning potential values to specific points in the circuit helps in calculating the currents through each branch. The potential across resistances can be expressed in terms of the assigned voltages, allowing for the application of Kirchhoff's current law. Using the superposition principle can also aid in determining the total current through each resistance by analyzing each voltage source separately. Understanding these methods is crucial for accurately solving parallel circuit problems.
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Homework Statement



[PLAIN]http://img600.imageshack.us/img600/7497/prob2.jpg

Homework Equations



I = V / R

The Attempt at a Solution



I tried to add the 3 voltages together but i didn't get a correct answer ..
I guess I'm missing something that has to be with the parallel Circuit or something like that ..
so how do we get the equivalent voltage ??

Thanks!
 
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welcome to pf!

hi hsnelect! welcome to pf! :wink:

you need to apply Kirchhoff's rules …

call the currents I2 I4 and I5

what do you get? :smile:
 
Assign a potential of 0 to the bottom right of the circuit.
Assign a Potential of V to the point where V1 and the 4 and 5 ohm resistances mee

work out the currents through the 3 branches of the circuit.
for example the potential across the 5 ohm resistance is V - V3 so the current through it is (V - V_3)/ 5

Now you can apply Kirchhof's current law, and then solve for V. The currents can
then be found with the 3 equations you found worked out above.

I don't think that adding the 3 voltages is going to get you anywhere. It seems you
might mean the the superposition principle: If you calculate the currents with only V1 in the circuit and the other voltage sources replaces by a wire, and calculate the currents through the resistances, and then you repeat this with V2 and V3, you can then sum the 3 currents you found for each resistance to get the current through that resistance with all the voltage sources present. I don't think that's an efficient method here.
 
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