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3 years ago

If the current in a circuit is 2 +j2 amps and the resistance is 3-j2 ohms, what is the voltage

Mathematics

1 answer:

victus00 [196]3 years ago

4 0

Answer:

The voltage is 10 + 2j volts.

Step-by-step explanation:

Ohm's first law:

By Ohm's first law, the voltage is the multiplication of the current by the resistance, that is:

$V = R*i$

The voltage is measured in Volts, the current in amps and the resistance in ohms.

In this question:

Both the current and the resistance have complex parts. We have to know that:

$j^{2} = -1$

We have that:

$R = 3 - 2j, i = 2 + 2j$

Then

$V = R*i$

$V = (3 - 2j)*(2 + 2j)$

$V = 6 + 6j - 4j - 4j^{2}$

$V = 6 + 2j - 4(-1)$

$V = 10 + 2j$

The voltage is 10 + 2j volts.

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