Question

Using the formula E=IZ, find the voltage E in a circuit when the current I is 3-j aamps and the impedance Z is 3+2j ohms

Answer 1

To find the voltage \( E \) using the formula \( E = IZ \), we need to perform the multiplication of the current \( I \) and the impedance \( Z \).

Given:
- Current \( I = 3 - j \) (in amps)
- Impedance \( Z = 3 + 2j \) (in ohms)

Now, we can perform the multiplication:

\[
E = I \cdot Z = (3 - j)(3 + 2j)
\]

Using the distributive property (also known as the FOIL method for binomials):

1. **First**: \( 3 \times 3 = 9 \)
2. **Outer**: \( 3 \times 2j = 6j \)
3. **Inner**: \( -j \times 3 = -3j \)
4. **Last**: \( -j \times 2j = -2j^2 \)

Now, recall that \( j^2 = -1 \), thus \( -2j^2 = 2 \). Now we combine all parts:

\[
E = 9 + 6j - 3j + 2
\]
Combine the real parts and the imaginary parts:

\[
E = (9 + 2) + (6j - 3j) = 11 + 3j
\]

Therefore, the voltage \( E \) in the circuit is:

\[
\boxed{11 + 3j} \text{ volts}
\]