Calculating the current of a circuit- resistance and volatge known, see picture

Click For Summary
SUMMARY

The discussion focuses on calculating the current flowing through three resistors (R(1) = 1Ω, R(2) = 2Ω, R(3) = 3Ω) with driving voltages V(1) = 1V and V(2) = 2V. The key equations used are Ohm's Law (V = RI) and the relationships between the currents through the resistors. The final calculated current through R(3) is I(3) = 1.37 A, although some participants noted potential errors in the derivation of the equations, particularly regarding the units in the final line of the calculations.

PREREQUISITES
  • Understanding of Ohm's Law (V = RI)
  • Basic circuit analysis techniques
  • Ability to manipulate algebraic equations
  • Familiarity with series and parallel resistor configurations
NEXT STEPS
  • Review circuit analysis methods for series and parallel resistors
  • Study the application of Kirchhoff's laws in circuit calculations
  • Learn about the implications of unit consistency in electrical equations
  • Explore advanced topics in circuit theory, such as Thevenin's and Norton's theorems
USEFUL FOR

Students studying electrical engineering, hobbyists working on circuit design, and anyone interested in understanding current flow in resistive circuits.

mmoadi
Messages
149
Reaction score
0

Homework Statement



What current flows through the resistors R(1) = 1Ω, R(2) = 2 Ω and R(3) = 3 Ω, if the driving voltages are V(1) = 1V and V(2) = 2V?

Picture:
http://item.slide.com/r/1/151/i/lvtwIwH-0j8A8o-XAKDtPRQO1eHAb4Ik/

Homework Equations



V= RI → I= V/ R

The Attempt at a Solution



V(1)= I(2)R(2) + I(1)R(1)
V(2)= I(3)R(3) – I(2)R(2)
I(1)= I(2) + I(3)

- From what I stated above, I conclude three equations:

1) V(1) = I(2)R(2) + [(I(2) + I(3))*R(1)]
2) V(2)= I(3)R(3) - I(2)R(2)
3) V(1)= [I(2)*( R(1) + R(2))] + I(3)R(1)

- I combined the three equations and simplify for the current that goes through ALL resistors, which in our case is I(3):

I(2)= (I(3)R(3) - V(2))/ R(2)
V(1)R(2)= [(I(3)R(3) – V(2))/ R(2))*(R(2) + R(1))] + I(3)R(1)R(2)
(V(1)R(2))/ (R(2) + R(1))= (I(3)R(3)) + [I(3)R(1)R(2))/ (R(2) + R(1))]
I(3)*[R(3) + (R(1)R(2)/ (R(2) + R(1))]= [V(1)R(2)/ (R(2) + R(1))] + V(2)

I(3)= [(V(1)R(2)/ (R(2) + R(1)) + V(2)] / [R(3)* (R(1)R(2)/ (R(2) + R(1))]
I(3)= 1.37 A

Are my calculations correct?
Thank you for helping!:smile:
 
Last edited by a moderator:
Physics news on Phys.org
mmoadi said:

Homework Statement



What current flows through the resistors R(1) = 1Ω, R(2) = 2 Ω and R(3) = 3 Ω, if the driving voltages are V(1) = 1V and V(2) = 2V?

Picture:
http://item.slide.com/r/1/151/i/lvtwIwH-0j8A8o-XAKDtPRQO1eHAb4Ik/

Homework Equations



V= RI → I= V/ R

The Attempt at a Solution



V(1)= I(2)R(2) + I(1)R(1)
V(2)= I(3)R(3) – I(2)R(2)
I(1)= I(2) + I(3)

- From what I stated above, I conclude three equations:

1) V(1) = I(2)R(2) + [(I(2) + I(3))*R(1)]
2) V(2)= I(3)R(3) - I(2)R(2)
3) V(1)= [I(2)*( R(1) + R(2))] + I(3)R(1)

Equation #3 here is the same as equation #1, just written slightly differently... so you really have only two independent equations. I'd just start from the original three you had above.

mmoadi said:
- I combined the three equations and simplify for the current that goes through ALL resistors, which in our case is I(3):

I(2)= (I(3)R(3) - V(2))/ R(2)
V(1)R(2)= [(I(3)R(3) – V(2))/ R(2))*(R(2) + R(1))] + I(3)R(1)R(2)
I don't see where that last line comes from. The units are off, so something must be wrong in there.
 
Last edited by a moderator:

Similar threads

  • · Replies 2 ·
Replies
2
Views
897
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K