# Incorrect current assumed in Hayt Ex 5.3

Here is Example 5.3 from Hayt "Engineering Circuit Analysis" p 129.

I am supposed to solve it using superposition principle. While doing it I incorrectly assumed direction of $$\i_x\$$. I assumed that arrow points from the right to the left. As far as I understand, my omission should not matter much. I am supposed to end up with answer of different sign, but the same magnitude. However, it did not happen, my solution below has completely different answer. I am staring at it for quite some time already, but cannot understand, why I obtained completely different solution.

My solution:

Since we use superposition principle, we do it in two steps. First step is to eliminate (open circuit) 3A current source. Everything related to this has $$\'\$$ sign to it. Second step is to short circuit 10 V voltage source. Everything related to the second step has $$\''\$$ sign to it.

Eliminating 3A source (open circuit). I ended up with the following equation (mesh approach):

$$\ 10 + 2 i^{'}_x + i^{'}_x - 2 i^{'}_x\$$ = 0

$$\i^{'}_x = -10\$$

Next we short circuit 10V source. Here I used node approach. First node is $$\v^{''}_1\$$ is at intersection of $$\2 \Omega\$$, $$\1 \Omega\$$ resistors and 3A current source. Second node $$\v^{''}_2\$$ is at intersection of $$\1 \Omega\$$ and $$\2 i^{''}_x\$$. Finally reference node is at the very bottom. Here are my equations:

$$\3 - \frac{v^{''}_1}{2}-\frac{v^{''}_1 - v^{''}_2}{1} = 0 \$$

one can notice that $$\ v^{''}_2 = 2 i^{''}_x\$$ and $$\\frac{v^{''}_{1}}{2} =i^{''}_{x}\$$

Hence:

$$\i^{''}_{x} = 3\$$

Adding two solutions we end up with $$\i_x = -7A\$$ as the answer

Correct solution from Hayt.

Elimination (open circuit) gives the following equation

$$\ -10 + 2 i^{'}_x + i^{'}_x + 2 i^{'}_x\$$ = 0

$$\i^{'}_x = 2A\$$

Short circuiting 10V voltage source gives the following equation

$$\\frac{v^{''}_1}{2}+\frac{v^{''}_1 - v^{''}_2}{1} = 3 \$$

$$\v^{''}_2 = 2i^{''}_x\$$

$$\v^{''}_1 = -2i^{''}_x\$$

Solving gives $$\i^{''}_x = -0.6A\$$

Adding two solutions we end up with $$\i^{}_x=1.4A\$$

If you swap the arrow direction for the current $$\i_x\$$ you also have to swap the sign on the dependent voltage source, and then you'll get the same answer with the opposite sign. For example, the first result (open current source) will be -2A rather than +10A.