Aneeka Hunt

2021-08-19

To solve:

The equation$E=IZ$ for given $E=57+67i,Z=9+5i$

The equation

davonliefI

Skilled2021-08-20Added 79 answers

1) Concept: Mathematical modeling is the application of the mathematical concepts in various field of study and also in daily practical side of finding solutions for various problems.

In Ohm's law$E=IZ$ any of the unknown value can be found provided the value of the others two.

i.e.$E=IZ$

$I=\frac{E}{Z}$

$Z=\frac{E}{I}$

2) Calculation:

Given,

$E=57+67i,Z=9+5i$

$E=IZ$

Dividing both sides on I

$\frac{E}{I}=\frac{IZ}{Z}$

$\frac{E}{Z}=I$

$I=\frac{E}{Z}$

On substituting the values of E and Z

$I=\frac{57+67i}{9+5i}$

Multiplying the conjugate$9-5i$ by the denominator with both the numerator and with the denominator

$=\frac{57+57i}{9+5i}\times \frac{9-5i}{9-5i}$

$=\frac{(57+67i)(9-5i)}{(9+5i)(9-5i)}$

Multiply each term of the first factor with each term of the factor in the numerator.

$=\frac{57\cdot 9+57\cdot (-5i)+\left(67i\right)\cdot 9+\left(67i\right)\cdot (-5i)}{(9+5i)(9-5i)}$

$=\frac{513-285i+603i-335{i}^{2}}{(9+5i)(9-5i)}$

Since${i}^{2}=-1$

$=\frac{513-285i+603i-335(-1)}{(9+5i)(9-5i)}$

$=\frac{513-285i+603i+335}{(9+5i)(9-5i)}$

Adding the real parts 513 and 335 together and the imaginary parts -285i and 603i together in the numerator

$=\frac{(513+335)+(-285i+603i)}{(9+5i)(9-5i)}$

Applying the rule for the product of sum and difference of two terms$(a+b)(a-b)={a}^{2}-{b}^{2}$

$=\frac{848+318i}{{9}^{2}-{\left(5i\right)}^{2}}$

$=\frac{848+318i}{81-25{i}^{2}}$

Since${i}^{2}=-1$

$=}">$

In Ohm's law

i.e.

2) Calculation:

Given,

Dividing both sides on I

On substituting the values of E and Z

Multiplying the conjugate

Multiply each term of the first factor with each term of the factor in the numerator.

Since

Adding the real parts 513 and 335 together and the imaginary parts -285i and 603i together in the numerator

Applying the rule for the product of sum and difference of two terms

Since

49

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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