cistG

2021-08-14

To solve:
The equation $E=IZ$ for given $I=10+4i,E=88+128i$

joshyoung05M

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,
$I=10+4i,E=88+128i$
$E=IZ$
Dividing both sides on I
$\frac{E}{I}=\frac{IZ}{Z}$
$\frac{E}{I}=Z$
$Z=\frac{E}{I}$
On substituting the values of E and I
$Z=\frac{88+128i}{10+4i}$
$=\frac{2\left(44+64i\right)}{2\left(5+2i\right)}$
$=\frac{44+64i}{5+2i}$
Multiplying the conjugate $5-2i$ by the denominator with both the numerator and with the denominator
$=\frac{44+64i}{5+2i}×\frac{5-2i}{5-2i}$
$=\frac{\left(44+64i\right)\left(5-2i\right)}{\left(5+2i\right)\left(5-2i\right)}$
Multiply each term of the first factor with each term of the factor in the numerator.
$=\frac{44\cdot 5+44\cdot \left(-2i\right)+64i\cdot 5+64i\cdot \left(-2i\right)}{\left(5+2i\right)\left(5-2i\right)}$
$=\frac{220-88i+320i-128{i}^{2}}{\left(5+2i\right)\left(5-2i\right)}$
$=\frac{220-88i+320i-128\left(-1\right)}{\left(5+2i\right)\left(5-2i\right)}$
$=\frac{220-88i+320i+128}{\left(5+2i\right)\left(5-2i\right)}$
Combining the real and imaginary parts
$=\frac{\left(220+128\right)+\left(320i-88i\right)}{\left(5+2i\right)\left(5-2i\right)}$
$=\frac{348+232i}{\left(5+2i\right)\left(5-2i\right)}$
Applying the rule for the product of sum and difference of two terms $\left(a+b\right)\left(a-b\right)={a}^{2}+{b}^{2}$

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