Voltage between two points of a circuit

[Gbox]

Homework Statement

In the circuit the battery is $10V$ and each resistor is $4\Omega$ find the voltage on AB

Homework Equations

1.KVL and KCL
2. $V=IR$

The Attempt at a Solution

1. $I_{1}=I_{2}+I_{3}$
   
2. $10-4I_{1}-4I_{1}-4I_{2}-4I_{1}=0\rightarrow 10-12I_{1}-4I_{2}=0$
3. $10-4I_{1}-4I_{1}-3*4I_{3}-4I_{1}=0\rightarrow 10-12I_{1}-12I_{3}=0$

$I_{1}=0.66$ $I_{2}=0.5$ $I_{3}=0.16$

Because the voltage at B is $0$ all that is left is to find the voltage on $A$ which is $10-8*0.66-4*0.16=4.08V$

Is it right?

 

[gneill]

Looks like you've rounded intermediate values and introduced significant errors into your significant figures. Also, I if I'm reading your final equation correctly, you've taking a "KVL walk" from B to A around the upper outside of the circuit, but haven't accounted for the 10 V source in the path.

 

[JaredMTg]

It looks like the second equation you wrote for Kirchhoff's law around the left-hand side loop has an error.

It should be 10 - 12 * (I1) - 4 * (I2) = 0, but you have a coefficient of 16 in front of I1.

Also, it looks like the terms in the third equation were also summed up incorrectly - the coefficient for both terms should be 12 (since three resistors in I1 and three resistors in I3).

 

[Gbox]

Sorry, wrong algebra, fixed and edited

 

[JaredMTg]

The equations are correct now. But as the other poster mentioned, your introduction of significant figures has caused your final answer to be imprecise.

The exact values for I1 is 0.666666... (i.e. 2/3), and the exact value for I3 is 0.1666666... (i.e. 1/6).

I would suggest using the fraction values for I1 and I3, or otherwise keeping more digits past the decimal point until you reach your final answer, and then round if applicable.

In that case, if you apply the equation you correctly wrote for the voltage between A and B, you will find that the answer is exactly 4 V (as opposed to your 4.08 V answer).