Question

the formula ( v = c(i) ) describes the relationship between voltage (v), current (c) and impedance (i). what is the voltage when the current in a circuit is ( 5 - 7i ) amps and the impedance is ( 4 + i ) ohms? (note, the variable for impedance is a capital i, but this does not mean that it is imaginary)

( 13 - 18i ) volts

( 13 - 23i ) volts

( 27 - 18i ) volts

( 27 - 23i ) volts

question 5 2 pts

which expression is equivalent to ( 1 - i )?

( i^{30}+i^{31} )

( i^{30}-i^{31} )

( i^{31}-i^{30} )

( (i^{30})(i^{31}) )

Explanation:

Step1: Substitute values into formula

Given \(C = 5 - 7i\) and \(I=4 + i\), substitute into \(V = C(I)\). So \(V=(5 - 7i)(4 + i)\).

Step2: Use FOIL method

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Step3: Simplify using \(i^{2}=-1\)

Since \(i^{2}=-1\), then \(V = 20-23i - 7\times(-1)=20 + 7-23i=27-23i\)

for Question 5:

Step1: Recall \(i^{n}\) cycle (\(i^{1}=i\), \(i^{2}=-1\), \(i^{3}=-i\), \(i^{4}=1\))

For \(i^{30}\), \(30\div4 = 7\cdots\cdots2\), so \(i^{30}=(i^{4})^{7}\times i^{2}=-1\).
For \(i^{31}\), \(31\div4 = 7\cdots\cdots3\), so \(i^{31}=(i^{4})^{7}\times i^{3}=-i\).

Step2: Calculate each option

• Option 1: \(i^{30}+i^{31}=-1 - i\)
• Option 2: \(i^{30}-i^{31}=-1-(-i)=-1 + i\)
• Option 3: \(i^{31}-i^{30}=-i-(-1)=1 - i\)
• Option 4: \((i^{30})(i^{31})=(-1)\times(-i)=i\)

Answer:

27 - 23i volts