Value of current through 12V battery

[damon669]

Homework Statement

a) Calculate the value of the 12V battery
b) Calculate the power dissipated in R_1,R_2 and R.

https://www.physicsforums.com/attachment.php?attachmentid=60819&stc=1&d=1376202905

Homework Equations

I_1+I_2+I_R=0

The Attempt at a Solution

For the left-hand circuit,
13=12-3I_R+1I_1

(2)
For the right-hand circuit,
14=12-3I_R+2I_2

(3)
Substituting for I_R in equation (2),

13=12-3(-I_1-I_2 )+I_1

13=12+4I_1+3I_2

1=4I_1+3I_2






(4)
Substituting for I_R in equation (3),

14=12-3(-I_1-I_2 )+2I_2

14=12+3I_1+5I_2

2=3I_1+5I_2






(5)
Multiplying equation (4) by 3 and equation (5) by 4 and subtracting to find I_2,

3=12I_1+9I_2

8=12I_1+20I_2

5=11I_2

I_2=5/11 A



Substituting for I_2 in equation (4) to find I_1,

1=4I_1+3(5/11)

I_1=-1/11 A






(9)
Substituting for I_1 and I_2 in (1) to find I_R,

I_R=-(-1/11)-(5/11)

=-4/11 A

=-0.3636 A

 

[NascentOxygen]

I was expecting the current through the 12V to be negative, since the other branches carry higher voltage batteries.

As a check, you could determine the voltage across each of the three parallel branches, and the voltages should all be equal.

To determine the power dissipated in each R, use P = I².R

 

[damon669]

For R1
P=I_1^2 R

=0.00826 W
For R2
P= I_2^2 R

=0.454*2

=0.90990 W
For R
P=I^2 R

=0.1322*3

=0.3967 W

 

[damon669]

so to check voltage is just

V=I/R

 

[NascentOxygen]

The voltage across the left-most branch is ¹/₁₁ x 1 + 13 = 13¹/₁₁ V

and a similar value should be found for the other branches.

 

[damon669]

left hand branch 1/11 x 1 +13 = 13.0909
right hand branch = 5/11 x 2 + 14 =14.0909

confused

 

[NascentOxygen]

left hand branch 1/11 x 1 +13 = 13.0909 ✔
right hand branch = 5/11 x 2 + 14 =14.0909 ✗

Their currents are in opposite directions. So figure out where you need a negative sign.

 

[damon669]

oh that was easy

-5/11 x 2 +14 = 13.0909

 

[damon669]

The voltage across the left-most branch is

1/11∙1+13=13 1/11 =13.0909 V

The voltage across the right-most branch is

-5/11∙2+14 =13.0909V

The voltage across the centre is

4/11∙3+12=13.0909V