Voltage Division - AC cct.

Thread Starter

Ammar i

Joined Feb 7, 2013
4
Hi,
We have this cut. :


and in the image below, I10 is calculated by using current division :



I tried to find it using voltage division but the answer is wrong :
Vx(of the upper node) = 100 * \(\frac{j5}{4+j5}\) = 60.98 + j48.78 V

=> I10 = \(\frac{Vx}{10 - j5}\) = 2.93 + 6.34 = 6.98 \(\ 65.2\)

which is wrong, WHY?

thanks
 
Last edited:

WBahn

Joined Mar 31, 2012
29,978
I tried to find it using voltage division but the answer is wrong :
Vx(of the upper node) = 100 * \(\frac{j5}{4+j5}\) = 60.98 + j48.78 V
What's your basis for this equation?

Before you say, "It's the formula for a voltage divider," think about what a voltage divider is and where the formula for a voltage divider comes from.
 
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