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Thread: Equivalent Impedance

  1. #1
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    Equivalent Impedance

    The equivalent impedance $\displaystyle Z$ of two impedances $\displaystyle Z_1$ and $\displaystyle Z_2$ in parallel is given by the the formula

    $\displaystyle \frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2}$

    If $\displaystyle Z_1 = 3 + j2$ and $\displaystyle Z_2 = 1 - j3$ calculate $\displaystyle Z$ giveing your answer in the form $\displaystyle r(\cos\theta + j\sin\theta)$ where $\displaystyle \theta$ is in radians.

    Any help with this would be highly appreciated.

    Thanks
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  2. #2
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    Hello anothernewbie
    Quote Originally Posted by anothernewbie View Post
    The equivalent impedance $\displaystyle Z$ of two impedances $\displaystyle Z_1$ and $\displaystyle Z_2$ in parallel is given by the the formula

    $\displaystyle \frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2}$

    If $\displaystyle Z_1 = 3 + j2$ and $\displaystyle Z_2 = 1 - j3$ calculate $\displaystyle Z$ giveing your answer in the form $\displaystyle r(\cos\theta + j\sin\theta)$ where $\displaystyle \theta$ is in radians.

    Any help with this would be highly appreciated.

    Thanks
    $\displaystyle \frac{1}{Z}= \frac{1}{3+2j}+\frac{1}{1-3j}$
    $\displaystyle =\frac{1-3j+3+2j}{(3+2j)(1-3j)}$

    $\displaystyle =\frac{4-j}{9-7j}$
    $\displaystyle \Rightarrow Z = \frac{9-7j}{4-j}$
    $\displaystyle =\frac{(9-7j)(4+j)}{4^2+1^2}$

    $\displaystyle =\frac{1}{17}(43-19j)$

    $\displaystyle =r(\cos\theta +j\sin\theta)$
    where
    $\displaystyle r\cos\theta = \frac{43}{17}$ and $\displaystyle r\sin\theta = -\frac{19}{17}$
    Square and add: $\displaystyle r^2(\cos^2\theta+\sin^2\theta) = \frac{43^2+19^2}{17^2}$

    $\displaystyle \Rightarrow r = 2.765$ (to 4 s.f.)

    Divide: $\displaystyle \tan\theta = -\frac{19}{43}$

    $\displaystyle \Rightarrow \theta = - 0.4161$ radians (to 4 s.f.)

    The method is sound, but check my working!

    Grandad
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  3. #3
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    thanks grandad!
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