enter image description hereSo I'm in an introductory circuit analysis class where all we have learned so far are Kirchhoff's and Ohm's Law. I am practicing how to solve these circuits, but for some reason I can't understand this example. I keep getting 3 Amps for \$i_5\$, but it says the answer is 2 A? I end up getting an equation like this:

$$3i_5 + 7i_5 - 2i_5 = 24$$

$$8i_5 = 24$$

$$i_5 = 3\ \mathrm{(V/Ohm)} = 3\ \mathrm{Amps}$$

  • \$\begingroup\$ Your polarity is wrong for v1. 3I5+7I5+2I5=24. Add the resistors in series and there's 24V across 12 ohms or 2A by inspection. \$\endgroup\$
    – John D
    Commented Feb 4, 2016 at 19:34
  • \$\begingroup\$ (1) What is the total resistance of the circuit? (series resistance) (2) what current would flow (Ohm's law) (= i5) \$\endgroup\$ Commented Feb 4, 2016 at 19:35

2 Answers 2


It looks like you got confused by the polarity of \$v_1\$. You can define a voltage to be whatever polarity you want, but that's just a notation thing. The physical polarity of the voltage across the component doesn't change. In the case of a resistor, the current always flows from higher voltage to lower voltage. This means \$v_1\$ is negative! Here's how the math works:

$$-v_1 + v_5 + v_2 = 24\ \mathrm V$$ $$v_1 = -i_5 \cdot 2\ \Omega$$ $$v_5 = i_5 \cdot 7\ \Omega$$ $$v_2 = i_5 \cdot 3\ \Omega$$

Note that both \$v_1\$'s value and the \$v_1\$ term in the KVL equation are negative. This gives us:

$$-(-2i_5) + 7i_5 + 3i_5 = 24$$ $$12i_5 = 24$$

So the answer is:

$$i_5 = 2\ \mathrm A$$

With practice, handling this sort of backwards polarity will become easy.


For calculating i5, simply use Ohm's law: V=IR. Voltage is 24V, total Resistance is 12 ohms. So you solve for current (current is the same if all elements are in series) using I=V/R I=24/12=2A. You are, unnecessarily, using Mesh analysis.

  • \$\begingroup\$ Thank you so much! I had to look up some of those terms. So why didn't the mesh analysis work? is it because the polarity in V1 was set in the wrong direction? \$\endgroup\$
    – Lee Jordan
    Commented Feb 4, 2016 at 19:40
  • \$\begingroup\$ Yes, regardless of the way the resistor polarity is drawn (+ - or - +) it is a voltage drop across the resistor, therefore you should have 2I5 not subtracted it. \$\endgroup\$
    – Mena Labib
    Commented Feb 4, 2016 at 19:44

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