To calculate the voltage $V$ of the circuit given the current $I$ and resistance $R$, we use the formula:

$V = I \cdot R$

Given:

• Current $I = 4 + j3$ amps
• Resistance $R = 5 - j3$ ohms

We perform the multiplication of these two complex numbers:

$(4 + j3) \cdot (5 - j3)$

We use the distributive property (FOIL method) for the multiplication:

$(4 + j3)(5 - j3) = 4 \cdot 5 + 4 \cdot (-j3) + j3 \cdot 5 + j3 \cdot (-j3)$

Now, calculate each term separately:

1. $4 \cdot 5 = 20$
2. $4 \cdot (-j3) = -j12$
3. $j3 \cdot 5 = j15$
4. $j3 \cdot (-j3) = -9$ (since $j^2 = -1$)

Combine these results:

$20 - j12 + j15 - 9$

Now, combine the real and imaginary parts:

• Real part: $20 - 9 = 11$
• Imaginary part: $-j12 + j15 = j3$

So, the voltage $V$ is:

$V = 11 + j3$

Thus, the voltage of the circuit is $11 + j3$ volts.

Do you want more details or have any questions?

Here are 5 questions you can ask next:

1. How do I perform operations on complex numbers?
2. Can you explain the concept of impedance in AC circuits?
3. How do I convert a complex number from rectangular to polar form?
4. What is the significance of the imaginary unit $j$ in electrical engineering?
5. Can you solve another problem involving complex numbers in electrical circuits?

Tip: When working with complex numbers in electrical engineering, always remember to separate real and imaginary parts for easier calculation and interpretation.