Calculating the currecnt of a circuit

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The discussion centers on calculating the current flowing through three resistors: R(1) = 1Ω, R(2) = 2Ω, and R(3) = 3Ω, with driving voltages V(1) = 1V and V(2) = 2V. The user derived three equations based on Ohm's Law (V = RI) and solved for the current I(3), concluding that I(3) = 1.37 A. The calculations were verified by another participant, who emphasized the need for consistency in the values of the currents across all equations.

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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:
 
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mmoadi said:

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)

I agree with these equations. Are your values for the currents consistent with them? You only gave I3 in your solution.
 

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