My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree2Thanks
  • 1 Post By weirddave
  • 1 Post By weirddave
Reply
 
LinkBack Thread Tools Display Modes
June 15th, 2016, 10:22 PM   #1
Banned Camp
 
Joined: May 2016
From: earth

Posts: 703
Thanks: 56

How to calculate current through Resistance

Hello,

in this circuit i applied KCL and KVL
i get these equation

I1 +I3 - I2=0 eq1 KCL

28 - 4*I1 - 2*I2=0 eq 2 KVL

7 - I3 - 2*I2 =0 eq 3 KVL

how to get individual value.
MMath is offline  
 
June 16th, 2016, 12:20 AM   #2
Senior Member
 
Joined: Apr 2014
From: UK

Posts: 965
Thanks: 342

Use substitution. The first equation can be rearranged as I2 = I1 + I3, so start with replacing I2 in equations 2 and 3. You will then be able to rearrange the 2nd equation and substitute it in the 3rd.
weirddave is offline  
June 16th, 2016, 01:44 AM   #3
Banned Camp
 
Joined: May 2016
From: earth

Posts: 703
Thanks: 56

I1 +I3 - I2=0 eq1 KCL

28 - 4*I1 - 2*I2=0 eq 2 KVL

7 - I3 - 2*I2 =0 eq 3 KVL

I1 +I3 = I2

28 - 4*I1 - 2*(I1+I3)
28- 4I1 - 2I1 -2I3

28 - 6I1 - 2I3 eq4


7 - I3 - 2*I2 =0
7-I3 - 2(I1+I3)
7-I3-2I1- 2I3
7-3I3 - 2I1 eq 5


28= 6I1 + 2I3
7=3I3 + 2I1

28= 6I1 + 2I3
7=2I1 + 3I3 * 3

28= 6I1 + 2I3
21=6I1 + 9I3

7=-7I3
I3=-1Amp


7 - I3 - 2*I2 =0
7 + 1 - 2*I2 =0
I2 = 4Amp

7 - I3 - 2*I2 =0
7 - I3 - 2*4 =0
7 -I3 -8=0
-1 -I3=0
I3=1amp


but in simulation answer is 5,1,4amps
MMath is offline  
June 16th, 2016, 03:43 AM   #4
Senior Member
 
Joined: Apr 2014
From: UK

Posts: 965
Thanks: 342

Equations 2 and 3 are wrong, some of the currents are drawn the opposite way to 'normal'
eq2:
28 - -4I1 - -2I2 = 28 + 4I1 + 2I2 = 0

eq3:
7 - -I3 - -2I2 = 7 + I3 + 2I2 = 0
weirddave is offline  
June 16th, 2016, 03:53 AM   #5
Banned Camp
 
Joined: May 2016
From: earth

Posts: 703
Thanks: 56

Quote:
Equations 2 and 3 are wrong, some of the currents are drawn the opposite way to 'normal'
i have taken current from the source to load as it do..

MMath is offline  
June 16th, 2016, 05:03 AM   #6
Senior Member
 
Joined: Apr 2014
From: UK

Posts: 965
Thanks: 342

The voltages in the left hand loop are:
28V - 0V = 28V
Voltage at [R1-] - voltage at [R1+] = --4I1 = 4I1
Voltage at [R2-] - voltage at [R2+] = 0 - - voltage at R2+ = 2I2

Note that the currents in R1 and R2 will be negative as measured in the direction shown.
Thanks from MMath
weirddave is offline  
June 16th, 2016, 05:42 AM   #7
Banned Camp
 
Joined: May 2016
From: earth

Posts: 703
Thanks: 56

i dont understand the current is from positive terminal to negative but here is difference.
MMath is offline  
June 16th, 2016, 06:21 AM   #8
Senior Member
 
Joined: Apr 2014
From: UK

Posts: 965
Thanks: 342

The arrows show the direction of measurement, either way around works, the sign just changes.
Maybe try a simpler circuit, take B2 and R3 out of the circuit and apply the rules to that (B1, R1 and R2 only).
Thanks from MMath
weirddave is offline  
June 16th, 2016, 06:47 AM   #9
Banned Camp
 
Joined: May 2016
From: earth

Posts: 703
Thanks: 56

Quote:
apply the rules to that (B1, R1 and R2 only).
then i can do.
it will behave like superposition theorem.
MMath is offline  
Reply

  My Math Forum > Science Forums > Physics

Tags
calculate, current, resistance



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Resistance as a function of Temperature dave daverson Algebra 1 October 28th, 2012 06:35 PM
Air resistance arron1990 Physics 5 March 28th, 2012 05:52 AM
Air resistance oswaler Physics 1 March 19th, 2008 07:36 PM
Resistance as a function of Temperature dave daverson Calculus 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.