What Is the Voltage Between Points a and b?

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SUMMARY

The voltage between points a and b in the given circuit can be calculated using Ohm's Law (V = IR). The circuit consists of resistors configured in series and parallel: two 5Ω resistors in series yield a total resistance of 10Ω, which is then combined in parallel with another 5Ω and a 10Ω resistor, resulting in an equivalent resistance of 2.5Ω. The total current from the 15V battery is 3A, leading to a voltage drop of 7.5V across the 2.5Ω resistor. Thus, the voltage between points a and b is 7.5V.

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Eman 5

Homework Statement


I want to know the voltage between a and b in the drawing below.[/B]
i1lnD.jpg


Homework Equations


V= IR

The Attempt at a Solution


The resistors 5Ω and 5Ω are in series:
(Req)1= 5+5=10Ω
The resistors (Req)1, 5Ω and 10Ω are in parallel:
1/(Req)2= 1/10+1/5+1/10=2/5
(Req)2=2.5Ω
The resistors (Req)2 and 2.5Ω are in series:
Rt=2.5+2.5=5Ω
I=V/Rt=15/5=3A
The voltage between a and b= IR
I have I but I don't know how to get R between the two points.
 

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Edited: Start by finding the total current put out by the battery. Then consider the voltage drop across the 2.5 Ω resistor.
 
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@Eman 5
You should know that iR, in this case, is the voltage drop between any branch in the circuit you drew, i.e. you can consider any branch of them to determine the voltage drop (potential difference) between the points a and b, because all branches in this circuit are connected to the points a and b.

For example, consider the above branch which has the battery and the 2.5 ohms resistance. Remember that the current passing through that branch is 3A.

Now determine the potential difference through that branch.
 
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The voltage across 2.5Ω resistor=IR=3×2.5=7.5V
The voltage drop across the above branch= 15−7.5=7.5V
Is this correct?
 
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Eman 5 said:
The voltage across 2.5Ω resistor=IR=3×2.5=7.5V
The voltage drop across the above branch= 15−7.5=7.5V
Is this correct?
Yes.
 
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