# ac power analysis, IrwinExt9.5

Discussion in 'Homework Help' started by PG1995, Jan 15, 2012.

1. ### PG1995 Thread Starter Well-Known Member

Apr 15, 2011
813
6
Hi

Please have a look on the attachment. Please help me with the query. Thank you.

Regards
PG

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Apr 15, 2011
813
6

3. ### Zazoo Member

Jul 27, 2011
114
43
Since one side of the branch is at ground (zero), the voltage at node A is the same as the voltage across the branch (19<72°V)

i.e. $V_{AG} = V_{A} - V_{G} = V_{A} - 0 = V_{A}$

4. ### PG1995 Thread Starter Well-Known Member

Apr 15, 2011
813
6
Thank you.

But suppose you don't know the voltage across the branch AG. In the attachment in my previous I stated how I found the voltage across AG by starting at G and then proceeding toward A. I'm just interested to know if it is possible to find the voltage across the branch AG by starting at A and proceeding toward G, the way I did in the attachment. Do you get me? Please let me know if it's possible. Thank you.

Best regards
PG

5. ### Zazoo Member

Jul 27, 2011
114
43
When you "travel" from G to A, you're basically completing a KVL loop that includes the unknown branch voltage (see the attached image):

So KVL for this loop is: $-25.4<45^{o} +12<0^{o} +V_{AG} = 0$

Applying the KVL loop in the opposite direction (A-to-G, clockwise) gives:
$-12<0^{o} +25.4<45^{o} -V_{AG} = 0$

The branch voltage is the same either way. You don't need to know the node voltage A (or even G)

• ###### KVL.jpg
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PG1995 likes this.
6. ### PG1995 Thread Starter Well-Known Member

Apr 15, 2011
813
6
Thanks a lot, Zazoo.

Best wishes
PG