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In an AC circuit, the voltage E, current I, and impedance Z are related by the
Formula E = I · Z
Find the current in a circuit with voltage 14 – 8j volts and Impedance is 2 – 3j ohms.
E = I · Z
Substitute E and Z values
14 – 8j = i · (2 – 3j)
Divide each side by (2 – 3j)
Multiply the numerator and denominator by the conjugate of the denominator
2 – 3j. Conjugate of 2 – 3j is 2 + 3j.
(Multiply the numerator using FOIL method) (Formula: )
Evaluate powers: .
(Multiply: )
(Multiply: )
Distribute terms using distributive property: .
(Subtract: )
(Substitute the imaginary unit value )
(Product of two same signs is positive)
Apply commutative property of addition: a + b = b + a.
(Add: 28 + 24 = 52 and 4 + 9 = 13)
(Write standard division form: )
(Divide: )
The current is 4 + 2j amps.
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