Equivalent Impedance

anothernewbie · #1 $Old$ November 3rd, 2009, 11:12 AM

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

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

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

Any help with this would be highly appreciated.

Thanks

Grandad · #2 $Old$ November 4th, 2009, 06:51 AM

Hello anothernewbie

Quote:

Originally Posted by anothernewbie $View Post$

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

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

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

Any help with this would be highly appreciated.

Thanks

$\frac{1}{Z}= \frac{1}{3+2j}+\frac{1}{1-3j}$

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

$=\frac{4-j}{9-7j}$

$\Rightarrow Z = \frac{9-7j}{4-j}$

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

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

$=r(\cos\theta +j\sin\theta)$

where

$r\cos\theta = \frac{43}{17}$ and $r\sin\theta = -\frac{19}{17}$

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

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

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

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

The method is sound, but check my working!

Grandad

srobrien · #3 $Old$ November 4th, 2009, 10:09 AM

thanks grandad!