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 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.
 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.
 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
 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
June 16th, 2016, 03:53 AM   #5
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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..

 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
 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.
 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
June 16th, 2016, 06:47 AM   #9
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 apply the rules to that (B1, R1 and R2 only).
then i can do.
it will behave like superposition theorem.

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