# Basic Circuit Analysis using Kirchhoff and Ohm's Laws

So 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:

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

$$8i_5 = 24$$

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

• 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. Commented Feb 4, 2016 at 19:34
• (1) What is the total resistance of the circuit? (series resistance) (2) what current would flow (Ohm's law) (= i5) Commented Feb 4, 2016 at 19:35

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:

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

$$i_5 = 2\ \mathrm A$$