Question

If the voltage in a circuit is 35 + 10i volts and the impedance is 4 + 4i ohms, what is the current (in amps)? Write your answer in the form a + bi.

Answer 1

Answer: 45-10i

----------
8

Step-by-step explanation:
You get the conjugate of 4+4i which is 4-4i, then you multiply the conjugate over the conjugate by the original problem, which looks like
35+10i 4-4i
---------- * ------------
4+4i 4-4i

After factoring these two equations you get

140-100i
-------------
32

You may simplify from here (no decimals for convenience)

resulting in

45-25i
-----------
8

Answer 2

The current in the circuit with a voltage of 35 + 10i volts and an impedance of 4 + 4i ohms is 3.125 - 3.125i amps.

Step-by-step explanation:

To find the current (in amps) when the voltage in a circuit is 35 + 10i volts and the impedance is 4 + 4i ohms, we use Ohm's law, which in the context of AC circuits with complex numbers is I = V / Z, where I is the current, V is the voltage, and Z is the impedance.

The calculation is as follows:

I = (35 + 10i) / (4 + 4i)

First, we must multiply the numerator and the denominator by the complex conjugate of the denominator to avoid division by a complex number, which yields:

I = (35 + 10i) * (4 - 4i) / ((4 + 4i) * (4 - 4i))

I = (140 - 140i + 40i - 40) / (16 + 16)

I = (100 - 100i) / 32

I = 3.125 - 3.125i

Therefore, the current I in the circuit is 3.125 - 3.125i amps, written in the form a + bi.