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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 7I3  2(I1+I3) 7I32I1 2I3 73I3  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  
Banned Camp Joined: May 2016 From: earth Posts: 703 Thanks: 56  Quote:
 
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. 
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). 
June 16th, 2016, 06:47 AM  #9  
Banned Camp Joined: May 2016 From: earth Posts: 703 Thanks: 56  Quote:
it will behave like superposition theorem.  

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calculate, current, resistance 
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