Aneeka Hunt

2021-08-19

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

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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
$=\frac{57+57i}{9+5i}×\frac{9-5i}{9-5i}$
$=\frac{\left(57+67i\right)\left(9-5i\right)}{\left(9+5i\right)\left(9-5i\right)}$
Multiply each term of the first factor with each term of the factor in the numerator.
$=\frac{57\cdot 9+57\cdot \left(-5i\right)+\left(67i\right)\cdot 9+\left(67i\right)\cdot \left(-5i\right)}{\left(9+5i\right)\left(9-5i\right)}$
$=\frac{513-285i+603i-335{i}^{2}}{\left(9+5i\right)\left(9-5i\right)}$
Since ${i}^{2}=-1$
$=\frac{513-285i+603i-335\left(-1\right)}{\left(9+5i\right)\left(9-5i\right)}$
$=\frac{513-285i+603i+335}{\left(9+5i\right)\left(9-5i\right)}$
Adding the real parts 513 and 335 together and the imaginary parts -285i and 603i together in the numerator
$=\frac{\left(513+335\right)+\left(-285i+603i\right)}{\left(9+5i\right)\left(9-5i\right)}$
Applying the rule for the product of sum and difference of two terms $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
$=\frac{848+318i}{{9}^{2}-{\left(5i\right)}^{2}}$
$=\frac{848+318i}{81-25{i}^{2}}$
Since ${i}^{2}=-1$

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