Value of current through 12V battery

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Discussion Overview

The discussion revolves around calculating the current through a 12V battery in a circuit with multiple branches and resistors. Participants explore the relationships between current, voltage, and power dissipation in the circuit components.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a series of equations to calculate the currents I_1, I_2, and I_R based on the circuit configuration.
  • Another participant expresses an expectation that the current through the 12V battery should be negative, suggesting that other branches with higher voltage batteries influence this outcome.
  • Power dissipated in resistors R1, R2, and R is calculated by participants using the formula P = I^2 R, with specific values provided for each resistor.
  • Participants discuss the voltage across each branch, indicating that the voltages should be equal across parallel branches.
  • Confusion arises regarding the calculations of voltages across branches, with one participant noting the need for a negative sign due to opposing current directions.
  • Subsequent posts clarify the voltage calculations for each branch, with some participants confirming their results while others express uncertainty.

Areas of Agreement / Disagreement

There is no consensus on the expected current through the 12V battery, with some participants suggesting it should be negative while others provide calculations that lead to different interpretations. The discussion remains unresolved regarding the voltage calculations across the branches, with conflicting results noted.

Contextual Notes

Participants rely on specific assumptions about the circuit configuration and the behavior of current in relation to voltage sources. There are unresolved mathematical steps and dependencies on the definitions of current direction and voltage across components.

Who May Find This Useful

This discussion may be useful for students and enthusiasts interested in circuit analysis, particularly those working on problems involving multiple voltage sources and resistors in parallel configurations.

damon669
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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
 

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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
 
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
 
so to check voltage is just

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

confused
 
damon669 said:
left hand branch 1/11 x 1 +13 = 13.0909 ✔[/size][/color]
right hand branch = 5/11 x 2 + 14 =14.0909 ✗[/size][/color]
Their currents are in opposite directions. So figure out where you need a negative sign.
 
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oh that was easy

-5/11 x 2 +14 = 13.0909
 
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
 

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