Magnitude/Direction of Current in a Circuit

  • Thread starter Thread starter vrobins1
  • Start date Start date
  • Tags Tags
    Circuit Current
Click For Summary
SUMMARY

The discussion focuses on calculating the magnitude and direction of current in a circuit with a battery voltage of B1 = 1.83 V. The user applied Kirchhoff's Loop and Junction Rules, resulting in the equations I1 + I2 - I3 = 0, -I3(56) - 5 - I1(22) = 0, and -I3(56) - 1.83 - I2(75) = 0. To solve for the unknown currents, the user is advised to substitute I3 with I1 + I2, leading to a system of two equations with two unknowns, which can be solved using algebraic methods or an online system solver.

PREREQUISITES
  • Understanding of Kirchhoff's Loop Rule
  • Familiarity with Kirchhoff's Junction Rule
  • Basic algebra for solving systems of equations
  • Knowledge of circuit components and their symbols
NEXT STEPS
  • Learn how to apply Kirchhoff's Laws in complex circuits
  • Study techniques for solving systems of linear equations
  • Explore circuit simulation tools like LTspice for practical applications
  • Review the significance of battery polarity in circuit analysis
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone seeking to understand circuit analysis using Kirchhoff's Laws.

vrobins1
Messages
21
Reaction score
0

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!
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
2K
Replies
22
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K