5
\$\begingroup\$

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

enter image description here

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:

enter image description here

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.

\$\endgroup\$
3
  • \$\begingroup\$ The mission is to find i_0 in the image below <-- is the mission to find it using mesh current methods? \$\endgroup\$
    – Andy aka
    Aug 5, 2022 at 7:19
  • \$\begingroup\$ @Andyaka No its just the method I used to solve the problem \$\endgroup\$
    – Moses Kim
    Aug 5, 2022 at 7:34
  • \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Aug 5, 2022 at 7:50

2 Answers 2

3
\$\begingroup\$

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:

enter image description here

\$\endgroup\$
3
\$\begingroup\$

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: -

enter image description here

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.