Mesh current method to find current on resistor

Question

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.

Answer 1

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:

Answer 2

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.