# Mesh current method to find current on resistor

The mission is to find i0 in the image below. The answer is 1.78 A.

I first combined the 1 and 4 Ω resistors to a 5 Ω resistor and combined the 3 and 6 Ω resistors to a 2 Ω resistor. Then I moved the current source to the side making a circuit like below:

From the left, there are four meshes. The first mesh has a clockwise mesh current of 5 A. I'll name the second mesh current i1, the third mesh current i2, both flowing clockwise. The fourth mesh current flows anti-clockwise with a current value of 3 A.

From the second and third mesh, we can get equations like below:

$$2(i_1-5)+5+7(i_1-i_2)=0 \\ 7(i_2-i_1)+5(i_2+3)=0$$

Solving the system of equations, we get:

$$i_1=-\frac{45}{59} \\ i_2=-\frac{100}{59}$$

This means that the current flowing in the 7 Ω resistor is:

$$i_o=\frac{55}{59}=0.93$$

Which is the wrong answer. What have I done wrong here? I checked the current in the other resistors using the result to validate the answer and it all added up to 8 A, so I assumed the answer I got was correct, but it wasn't.

• The mission is to find i_0 in the image below <-- is the mission to find it using mesh current methods? Commented Aug 5, 2022 at 7:19
• @Andyaka No its just the method I used to solve the problem Commented Aug 5, 2022 at 7:34
• @MosesKim I don't get your numbers using two equations I wrote for the same two loop currents you used. But I want to know what you believe is the correct value for that resistor.
– jonk
Commented Aug 5, 2022 at 7:50

Your approach is sound. You just didn't write out the equations correctly. Assuming clockwise current loops and generating the algebra in that direction:

\begin{align*} 0\:\text{V}-\left(i_1-5\:\text{A}\right)R_2+5\:\text{V}-\left(i_1-i_2\right)R_3&=0\:\text{V} \\\\ 0\:\text{V}-\left(i_2-i_1\right)R_3-\left(i_2+3\:\text{A}\right)R_5&=0\:\text{V} \end{align*}

Note that you add the $$\3\:\text{A}\$$ current because, as you have accepted it to be, it flows in the same direction in $$\R_5\$$ as does the clockwise $$\i_2\$$. So they sum up.

These solve out as $$\i_1=\frac{75}{59}\:\text{A}\$$ and $$\i_2=-\frac{30}{59}\:\text{A}\$$. The current in $$\R_3\$$ is $$\i_1-i_2=\frac{75}{59}\:\text{A}-\left(-\frac{30}{59}\:\text{A}\right)\$$ and that's $$\\frac{105}{59}\:\text{A}\$$ or $$\1.779661\:\text{A}\$$.

LTspice agrees:

The mission is to find i0 in the image below. The answer is 1.78 A.

You can solve it using circuit reduction techniques like this: -

And you are left with a 15 volt source driving 2 Ω || 5 Ω in series with 7 Ω.

The current is 15 volts divided by 8.42857 Ω = 1.77966 amps.