Sure, let's work through this problem step-by-step.
1. Identify the given current [tex]\(I\)[/tex]:
[tex]\[
I = 3 + 2
\][/tex]
Here, the current [tex]\(I\)[/tex] is initially given as [tex]\(3 + 2\)[/tex].
2. Calculate the resistance [tex]\(Z\)[/tex] using the given relationship:
[tex]\[
Z = 2 - I
\][/tex]
Substituting the given value of [tex]\(I\)[/tex]:
[tex]\[
Z = 2 - (3 + 2) = 2 - 5 = -3
\][/tex]
So, the resistance [tex]\(Z\)[/tex] is [tex]\(-3\)[/tex].
3. Calculate the voltage [tex]\(E\)[/tex] using the formula [tex]\(E = I \cdot Z\)[/tex]:
Substitute the calculated values of [tex]\(I\)[/tex] and [tex]\(Z\)[/tex]:
[tex]\[
E = (3 + 2) \cdot (-3)
\][/tex]
4. Express [tex]\(I\)[/tex] in terms of its real and imaginary parts (since we're given [tex]\(I\)[/tex] to be [tex]\(3 + 2\)[/tex], where 2 is understood to be the imaginary part here):
[tex]\[
I = 5 \text{ (no imaginary part since it simplifies to a real number)}
\][/tex]
5. Calculate the voltage [tex]\(E\)[/tex], now as the product of real and imaginary parts:
For the real part:
[tex]\[
E_{\text{real}} = 5 \cdot (-3) = -15
\][/tex]
For the imaginary part:
[tex]\[
E_{\text{imag}} = 0 \text{ (since there is no imaginary part in \(I\))}
\][/tex]
6. Combine the results:
[tex]\[
E = -15 + 0j
\][/tex]
Thus, the voltage [tex]\(E\)[/tex] is [tex]\(-15 + 0j\)[/tex], which is simply [tex]\(-15\)[/tex] when considering only the real number scenario. No imaginary part is present here.
So, the answer to the question is:
[tex]\[
\text{The voltage of the circuit is } -15
\][/tex]
