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So I thought I knew what I was doing but, this 3rd voltage source is totally throwing me off. Any explanation is very much welcome.

schematic

simulate this circuit – Schematic created using CircuitLab

So I have

I1 = I2 + I3 ------------------- (1)

-I1(3) - 30 - I1(2) = 0 -------- (2)

-I2(4) - 10 - I2(5) + 30 = 0 --- (3)

Solve for I1 in (2), we get:
-(5)I1 - 30 = 0 I1 = -6

Sub I1 = - 6 into (1) and then that into (3)

(I2 + 6)(4) - 10 + (I2 + 6)(5) + 30 (9)I2 + 74 = 0 I2 = - 8.2

Therefore I1 = -(-8.2) - 6 = 2.2

Does this look right? I want to be able to double check my answer using power but am unsure what voltage to use for each resistor. Any help/explanation is greatly appreciated.

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  • \$\begingroup\$ I3 is through a short circuit. Your analysis is invalid. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 16, 2015 at 4:28
  • \$\begingroup\$ Try applying KVL then use KCL to find I3. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Oct 16, 2015 at 5:13
  • \$\begingroup\$ @IgnacioVazquez-Abrams could you provide a bit more detail why? \$\endgroup\$
    – Rio
    Commented Oct 16, 2015 at 20:08
  • \$\begingroup\$ Because voltage sources other than the one being analyzed are considered short circuits. And current sources are considered open circuits. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 16, 2015 at 21:23
  • \$\begingroup\$ @IgnacioVazquez-Abrams so I should ignore any other power source when analyzing a loop? \$\endgroup\$
    – Rio
    Commented Oct 17, 2015 at 3:25

3 Answers 3

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You should try something different, using meshes. I'll do the one on the left.

$$ +20-3I_l-30-3I_l=0 \implies I_l = -\frac{5}{3}\ A $$

On a side note, this problem could be easily solved with superposition. If you know about it, then you might want to give it a shot.

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  • \$\begingroup\$ Thanks! I realize from your equation that I forgot to include the +20 in my 2nd equation. I am required for this particular problem to use Kirchhoffs law. \$\endgroup\$
    – Rio
    Commented Oct 16, 2015 at 20:11
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First calculate the current through the chosen resistor by picking one voltage source and by considering the rest closed connections. Do this for all voltage sources so that every time you have one active source. Add these current up and there you have the total current through that one resistor.

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  • \$\begingroup\$ One tip: Redraw after choosing one source, the setup of the design can really be confusing sometimes, although it is very simple. \$\endgroup\$
    – Weaverworm
    Commented Oct 16, 2015 at 9:31
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The current flowing downward across the 10 v battery is (20/9)A.

The current flowing downward across the 20 v battery is (10/6)A.

The current flaowing upward across the 30 v battery is (35/9)A.

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  • \$\begingroup\$ Please don't hand out solutions to obvious homework questions, even old ones. \$\endgroup\$
    – Elliot Alderson
    Commented Apr 26, 2023 at 19:25

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