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I am confused about when to use KVL/KCL or ohm's law to solve the circuit question?

I need to compute voltage in point B and point C, every current pass through R1 R2 R3 R4

enter image description here

R1= 100 ohm, R2=300 ohm, R3=250 ohm, R4=500 ohm

V1= 5V, Is=20mA

I tried to use ohm's law to find all current first, but I don't know how to use Is to help me. Or find out the total resistance then finding total current?

And how to use Is when applying KCL, is it a current source?

Or I can separate circuit into two independent circuits to solve the question.

I hope someone can help me to clear my electricity concept.


the resistance of r1 and r2 =1/(1/100+1/250)= 500/7 ohm ≈ 71.4 ohm

total resistance of a circuit = 500/7 + 300 + 500 = 6100/7 ohm ≈ 871 ohm

current supply from 5V : I = V/R = 5/(6100/7) = 7/1220 A ≈ 5.74mA

I1=5.73 mA

I1=I2+I3

5.73= (VB-VGround)/250 + (VB-VC)/300 = VB/250 + (VB-VC)/300

5.73=VB/250 + (VB-VC)/300 -(1)

I4+Is = (VC-VGround)/500

I4+ 20mA = VC/500 -(2)

apply KVL: VB-(VB-VC)+VC = 0 -(3)

Is it correct, should I continue?

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  • \$\begingroup\$ I mean, no offense, I see you are a new user, but I cannot see any effort from you, at least attempting to solve the problem. \$\endgroup\$ – Atizs Oct 10 '18 at 12:59
  • \$\begingroup\$ You say that you have tried to use Ohm's law etc. If you show what work you have done, and the steps you have taken, then we can help you figure out when and how you got confused and help you get back on the right track \$\endgroup\$ – MCG Oct 10 '18 at 13:06
  • \$\begingroup\$ I apologize for my lazy and naive, I will show my step. \$\endgroup\$ – PAK KIN HON Oct 10 '18 at 15:51
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When in doubt simplify: -

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And if you don't understand what I've done look up Thevenin's and Norton's theorums about replacing current sources with voltage sources and vice versa.

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  • \$\begingroup\$ Good start for the user. \$\endgroup\$ – mike65535 Oct 10 '18 at 14:48
  • \$\begingroup\$ But how does 6V come out? Why can you add a voltage source between R2 and R3? \$\endgroup\$ – PAK KIN HON Oct 10 '18 at 16:21
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    \$\begingroup\$ A 20 mA current source in parallel with 300 ohms is identical to a 6 volt source in series with 300 ohms. Both have the same short circuit current (20 mA) and open circuit voltage (6 volts). \$\endgroup\$ – Andy aka Oct 10 '18 at 16:24
  • \$\begingroup\$ Ok, so how do you know the current supplied from 5V? You don't need to calculate the total resistance first? By V=IR \$\endgroup\$ – PAK KIN HON Oct 10 '18 at 16:38
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    \$\begingroup\$ The 5 volt source is exclusively in series with 100 ohm therefore it is equivalent to a 50 mA current source in parallel with 100 ohms. Wikipedia has a very good page on the theorems I mentioned. \$\endgroup\$ – Andy aka Oct 10 '18 at 18:18

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