the current in a circuit is 2 minus j 5 amps and the resistance is 1 plus j 3 ohms, what is the voltage?


A.)
17 plus j 11 volts


B.)
13 plus j volts


C.)
17 plus j volts


D.)
13 plus j 11 volts

To calculate the voltage across a circuit, we use Ohm's Law, which states:

\[
V = I \times R
\]

Where:
- \( V \) is the voltage,
- \( I \) is the current, and
- \( R \) is the resistance.

Given:
- \( I = 2 - j5 \) amps,
- \( R = 1 + j3 \) ohms.

Now, multiply the current \( I \) by the resistance \( R \):

\[
V = (2 - j5) \times (1 + j3)
\]

Using the distributive property (FOIL method):

\[
V = (2 \times 1) + (2 \times j3) + (-j5 \times 1) + (-j5 \times j3)
\]
\[
V = 2 + j6 - j5 + 15
\]

Simplify the real and imaginary parts:

\[
V = (2 + 15) + (j6 - j5)
\]
\[
V = 17 + j1 \text{ volts}
\]

Thus, the voltage is:

\[
V = 17 + j \text{ volts}
\]

The correct answer is **C.) 17 plus j volts**.

Would you like further details or explanations?

Here are 5 related questions:
1. How is the distributive property applied in complex number multiplication?
2. What does the imaginary part of a complex number in a circuit represent?
3. How do you convert polar form to rectangular form in AC circuit analysis?
4. Can Ohm's law be applied directly to AC circuits with impedance?
5. How would the voltage change if the current were purely real?

**Tip:** Always simplify complex numbers by separating real and imaginary components after multiplication.

The current in a circuit is 2 minus j 5 amps and the resistance is 1 plus j 3 ohms, what is the voltage?

Learn how to calculate the voltage in an AC circuit using complex numbers. Given the current 2 - j5 amps and resistance 1 + j3 ohms, this problem demonstrates the application of Ohm's Law in AC circuits. Suitable for undergraduate students in electrical engineering.

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