Solving a circuit with 5 resistors using Kirchhoff's laws.

Thread Starter

mdalala

Joined Dec 2, 2018
4
Hello everyone.. I've been trying to figure out for 2 days how to solve this circuit.. I attach a picture below:

https://ibb.co/59qNmKx

I attach a link because for some reason the picture doesn't work..


R1 = 15 ohms
R2 = 22 ohms
R3 = 10 ohms
R4 = 15 ohms
R5 = 15 ohms
E1=E2=35 V

I've applied Kirchhoff's laws and got these equations:

15I2+10I6 = 35
22I5-15I2+15I3+15I3 = 0
-10I6-22I5 = 35
I1 = I2 + I3
I3 = I5+I4
I6 = I2+I5
I1 = I6+I4

from which I got that I3 = 2.33 A but I am stuck at that and cannot solve any other currents.. Are my equations right ? Do I understand correctly that I have to define junction points myself and that there will be 6 currents because junction points are not given ? Thanks for help...
 

dl324

Joined Mar 30, 2015
11,232
Welcome to AAC!

Circuit in question.
upload_2018-12-2_7-53-40.png

Please redraw with component values on the schematic. Is the polarity of E1 correct?
 

dl324

Joined Mar 30, 2015
11,232
Circuit in question:
upload_2018-12-2_8-16-18.png

Have you studied anything that lets you solve networks with multiple independent power sources?
 

WBahn

Joined Mar 31, 2012
26,047
Hello everyone.. I've been trying to figure out for 2 days how to solve this circuit.. I attach a picture below:

https://ibb.co/59qNmKx

I attach a link because for some reason the picture doesn't work..


R1 = 15 ohms
R2 = 22 ohms
R3 = 10 ohms
R4 = 15 ohms
R5 = 15 ohms
E1=E2=35 V

I've applied Kirchhoff's laws and got these equations:

15I2+10I6 = 35
22I5-15I2+15I3+15I3 = 0
-10I6-22I5 = 35
I1 = I2 + I3
I3 = I5+I4
I6 = I2+I5
I1 = I6+I4

from which I got that I3 = 2.33 A but I am stuck at that and cannot solve any other currents.. Are my equations right ? Do I understand correctly that I have to define junction points myself and that there will be 6 currents because junction points are not given ? Thanks for help...
It's generally better to use the symbolic values as long as possible. Makes error detection much easier.

So your equations would be

R4·I2 + R3·I6 + 35 V = 0

Notice that as you go counterclockwise around the left-hand loop, you encounter voltage DROPS across all three components. So that's one mistake you made.

R2·I5 - R4·I2 + R1·I3 + R5·I3 = 0

This one looks fine.

-R3·I6 - R2·I5 + 35 V = 0

So this has the same mistake as your first one.

Unless... Are you using electron current or conventional current? The fact that you draw I1 and I4 as coming out of the bottom of the batteries implies you might be using electron current (which is unfortunate).
 

Thread Starter

mdalala

Joined Dec 2, 2018
4
It's generally better to use the symbolic values as long as possible. Makes error detection much easier.

So your equations would be

R4·I2 + R3·I6 + 35 V = 0

Notice that as you go counterclockwise around the left-hand loop, you encounter voltage DROPS across all three components. So that's one mistake you made.

R2·I5 - R4·I2 + R1·I3 + R5·I3 = 0

This one looks fine.

-R3·I6 - R2·I5 + 35 V = 0

So this has the same mistake as your first one.

Unless... Are you using electron current or conventional current? The fact that you draw I1 and I4 as coming out of the bottom of the batteries implies you might be using electron current (which is unfortunate).
Thank you.
 
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