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Elementary resistive network

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mabauti

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I want to know the voltage VA on this circuit :
 

By simple KVL, it must be

\[(V_1-V_2)\frac{R_3}{R_1+R_3}\]
 

Tell me if I'm wrong :


let's assume that V1 = 0, V2 = 5, R3 =5, R2=10 and R1=10
so the equivalent circuit (R1 will be in parallel with R2) will be in series with R3 and RT = R3 + (R1||R2) = 5 + 5 ; so VA = V2/2 => VA =5/2 =2.5

the formula VA = (V1-V2 ) * R3 /(R3 + R1) produces : -5 *5/(5+10) = -1.66 :?:

Added after 34 minutes:

Never mind I just got the solution. Thx anyway.
 

Hi, jayc, your formula cannot be correct since it ignores R2: R2=0 should lead to VA=0.

mabauti, solve by applying superposition not by applying KVL :)
 

Hi

The volatage in a is

Va=R2*(V1/(R1+R2) + V2/(R2+R3))

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Sal
 

As the network is linear you can apply superposition principle.

1) set V2 to 0 (Short V2 generator) and derive voltages and currents in the circuit

2) set V1 to 0 (Short V1 generator) and derive voltages and currents in the circuit

Now summing the results obtained at point 1 and 2 you will obtain the voltages and currents when both generators are on.
 

mabauti,
Millman had you in mind when he developed "Millman's Theorem". It covers exactly the kind of problem you describe, except it is more general in that it covers networks with an arbitrary number of voltage sources. See, for example
.
h**p://www.allaboutcircuits.com/vol_1/chpt_10/5.html
.
Regards,
Kral
 

hi..u can do this by
1. by using superposition
2. by using kvl and ohms law
1. in supersition first set v1=0 and find VA.
next set v2=0 and find vA.then add both VA values to get total vaue. b careful with current directions.
2. u can apply kvl for path R1,mid resi. while using kvl curr thro mid resi should b i1+i2.
 

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