Magnitude/Direction of Current in a Circuit

[vrobins1]

Homework Statement

Find the magnitude of the current in each branch of the circuit shown below, in which B1 = 1.83 V. Specify the direction of each current.

http://img183.imageshack.us/img183/6951/physics.gif

I labeled + and - and drew the assumed currents, with I₃ flowing to the right along the middle, I₁ flowing CW about the top loop, and I₂ flowing CCW about the bottom loop.

Homework Equations

The Loop and Junction Rules apply.

The Attempt at a Solution

I applied the junction rule first.

I₁ + I₂ - I₃ = 0
I₁ + I₂ = I₃

Then I applied the loop rule to the top loop.

-I₃(56) - 5 - I₁(22) = 0

And then I applied the loop rule to the bottom loop.

-I₃(56) - 1.83 - I₂(75) = 0

I know I have to use these equations to solve for the unknown I's, but I am lost on what to substitute/do next. Any insight would be great. Thanks!

 

[Delphi51]

Use I1 + I2 = I3 to replace both of the I3's in the other equations with I1 + I2.
Then you will have a system of two equations with two unknowns.
You can solve one of them for I2, then sub that into the other one to get one equation with one unknown. Alternatively, use the system solver at http://www.analyzemath.com/Calculators/Calculator_syst_eq.html to deal with the system of 2 equations.

What is the polarity on the battery symbol? Years ago they had it wrong - the small end was negative. Have they corrected that in your class or is it still that way? Of course it affects your answer so you need to know which end is positive.

 

[nlightner]

I would approach this problem by first analyzing the circuit and identifying the relationships between the different components. It appears that the circuit is a combination of series and parallel connections, with a voltage source (B1) and three resistors (5Ω, 22Ω, and 75Ω).

To find the magnitude of the current in each branch, we can use Ohm's Law (V=IR) and the concept of Kirchhoff's Laws. The junction rule states that the sum of the currents entering a junction must equal the sum of the currents leaving the junction. Using this rule, we can set up the following equation:

I1 + I2 = I3

Next, we can apply the loop rule to the top loop and bottom loop. The loop rule states that the sum of the potential differences around a closed loop must equal zero. Using this rule, we can set up the following equations:

-5 - 22I1 - 56I3 = 0
-1.83 - 75I2 - 56I3 = 0

We now have three equations and three unknowns (I1, I2, and I3). We can use algebraic methods to solve for these unknowns. One approach would be to solve for one variable in terms of the others and then substitute that into the other equations.

For example, we can solve for I1 in terms of I2 and I3 from the first equation:

I1 = I3 - I2

We can then substitute this expression for I1 into the second and third equations. After some algebraic manipulation, we can solve for I2 and I3.

Once we have solved for I1, I2, and I3, we can use Ohm's Law to find the magnitude of the current in each branch. For example, the magnitude of the current in the top branch (I1) would be:

I1 = (V1 - V2)/R1 = (5V - 0V)/22Ω = 0.227 A

The direction of the current can be determined based on the assumed direction in the problem statement. In this case, I1 is flowing clockwise about the top loop, I2 is flowing counterclockwise about the bottom loop, and I3 is flowing to the right along the middle branch.

In summary, to find the magnitude and direction of the current in each branch