Help me with solving a Kirchhoffs law task

[bantamm]

kirchhoff's law help

it is task 1 which pains me so.

I have completed the assignment, but am now told that task1 b & c are incorrect, and i cant figure out why.

i used the equations...

total current = 13/3.5 = 3.174

1ohm resistor = ItR1 = 3.714 x 1
0.5ohm resistor = ItR0.5 = 3.714 x 0.5
2 ohm resistor = ItR2 = 3.714 x 2

the answers i have, im told are wrong.

i'm now told that i have to use the equation E1-E2= IR1+IR2+IR3 for task 1b.


can you guys help?? this is driving me mad, as i thought iwas correct, and i need to get this assignment out of the way.

 

[Roshdy]

Re: kirchhoff's law help

Your mistake is
you assumed that the sum of the voltages is aplied to the sum of the resistors which is not the case, ach loop has to be calculated separatly by kirchhoff's law.

 

[bangash]

Re: kirchhoff's law help

u have two loops. u need to use MESH analysis. check for currents from both voltage sources and then try it. use same procedure fo 8V source also. and then add currents for central resistor if direction of flow of current is same.

 

[bantamm]

kirchhoff's law help

thanks for the help.

i realised what i had done wrong, and re- calculated, now the answer is correct.

 

[dspti]

Re: kirchhoff's law help

Can u show the correct answer?

I think
a) I = 4/3
b) L=7/3V M=2/3 R=16/3

 

[bantamm]

kirchhoff's law help

it seems i could have been wrong for task 1c. the power dissipated.

i used the equation P=VI

should i have used P=I(squared)R ???

 

[Miguel Gaspar]

Re: kirchhoff's law help

You can use the superimposition theorem.
First for the first source 5V there is a circuit with

5v in series with the 1Ω, and both in paralell with .5||2 = 0.4Ω

The current due 5V Source = 5V/1.4 = 3.57 A

Voltage in the 1Ω = 1Ω *3.57A = 3.57V

Voltage in the 0.5Ω resistor = 5-3.57 = 1.43V

Current in the 0.5Ω resistor = 1.43v /0.5Ω = 2.86A

Current in the 2Ω resistor = 3.57A - 2.86A = 0.71A

Second for the source 8V there is a circuit with

1Ω ||0.5Ω = 0.33Ω

so i = 8/(2+0.33) = 3.43A

V in 2Ω = 3.43A* 2Ω = 6.86V

V in the other resistors = 8 - 6.86 = 1.14V

V IN 0.5Ω = 1.14/0.5 = 2.28A


Finnaly Vin 0.5Ω = 2.86A + 2.28A

and you can go on the same way to answear the rest of the questions