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I can only solve this using Kirchhoff Laws and Ohm's Law.

The first thing i tried was finding the current through R3 , Ir3=1mA (Ir3 = 12/12.10^3 = 1mA).

Then using KCL on node A: Ir2 = Ir3 + Ir1 <=> 2 = 1 + Ir1 <=> Ir1 = 1mA.

And finally using KVL to find Vs : -Vs+R1*Ir1+R2*(Ir1-Ir3)=0<=>-Vs+6+6(1-1)=0<=>-Vs =-6<=> Vs=6.(Which is wrong)

But i can't get the correct answer, i suspect im doing wrong the KVL and i've been trying for the last hour different things . Appreciate any help...(I know the correct answer ,which is 24V, in my question i expected someone to point where in my logic it went wrong to get a different answer)

  • 1
    \$\begingroup\$ I would first calculate the voltage across R2, since you are given its resistance, and the current through it. \$\endgroup\$ – Peter Bennett Oct 28 '17 at 15:38
  • \$\begingroup\$ @PeterBennett done that, its 12V but i can't see how to apply that information...besides using it as the R2*(Ir1-Ir3), which will still give a wrong answer \$\endgroup\$ – Alex Oct 28 '17 at 15:38
  • \$\begingroup\$ If there is 12 volts across R2, what is thje current through R3? \$\endgroup\$ – Peter Bennett Oct 28 '17 at 16:17
  • \$\begingroup\$ "But I can't get the correct answer" - What is the 'correct' answer then - you indicate you know it but don't state it or is it the case you don't know it but can't get an answer using your working. \$\endgroup\$ – JIm Dearden Oct 28 '17 at 16:17
  • \$\begingroup\$ @JImDearden the answer supposedly is 24V i dont know how to get there. \$\endgroup\$ – Alex Oct 28 '17 at 16:24

Here's what you should do:

  • Calculate the voltage at node A, given the current through \$R_2\$ is known
  • Having the voltage at node A, find the current through \$R_3\$
  • Find the current through \$R_1\$ by applying KCL at node A
  • Now use KVL to figure out \$V_S\$

Throughout all of this stick to one convention for current/voltage sign for resistors and voltage sources. For example, here's one convention that people usually use:

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  • \$\begingroup\$ Is the 2mA current going upward or downward? Also pick the ground node at the negative side of voltage sources. If the 2mA current is flowing downward then the voltage at node A will be +12V with respect to the ground. \$\endgroup\$ – dirac16 Oct 28 '17 at 16:27
  • \$\begingroup\$ So now applying KCL at node A gives you 12V across \$R_2\$, since \$\frac{V_A-0V}{6k}=2mA\$, or \$V_A=12V.\$ This also means that the current through \$R_3\$ is zero. \$\endgroup\$ – dirac16 Oct 28 '17 at 16:32
  • \$\begingroup\$ No current will flow through \$R_3\$ because there is no potential difference across it, hence the current through \$R_3\$ will be 0A. \$\endgroup\$ – dirac16 Oct 28 '17 at 16:36
  • \$\begingroup\$ What do you mean there is no potential difference? Isn't the source giving 12V? \$\endgroup\$ – Alex Oct 28 '17 at 16:37
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    \$\begingroup\$ Yes, correct. The voltage across \$R_3\$ is 0V because both of its ends are held at +12V so no current flows through it. \$\endgroup\$ – dirac16 Oct 28 '17 at 16:46

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