Source exchange help

Thread Starter

luck

Joined Jan 25, 2008
8
Attached (8_96.JPG )is a problem which I am suppose to solve for V using source exchange.

What I have tried:
I combined the 2 A source with the 1 Ohm resister to a 2 V 1 Ohm Thevenin As seen in the second attached image (8_96_2.JPG ). I then transformed it as seen in attachment 3 (8_96_3.JPG ). I am not positive I can do this but I think I can. From here it is just a voltage divider so it would be 20*2/(3-1J) which gives me 12.65<18.43˚

The answer is suppose to be 10.23<43.82˚

Please let me know what I am doing wrong, Thanks
 

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Ron H

Joined Apr 14, 2005
7,014
In your second schematic, where you did the Thevinin equivalence, The 1 ohm resistor needs to be in series with the 2V source, not in series with the capacitor.
 
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