AC mesh analysis and power calculations

Discussion in 'Homework Help' started by regexp, Jan 2, 2011.

1. regexp Thread Starter New Member

Nov 20, 2010
24
0
Hi,

In this circuit.

I need to determine the power dissipated in the resistance and also the reactive power delivered to the source V1.

Starting by writing the mesh equations for I1:

$8I_{1} -I_{1}\cdot J2 + I_{2}\cdot J2 = 40$

$I_{2}J2 -I_{1}J2 +I_{2}J4 = -j20$

Subtracting the second from the first gives
$8I_{1} -I_{2}J4 = 40+j20$

The correct answer is supposed to be $8I_{1}-J4I_{1} = 40+J20$

I don't really see where i went wrong =/

2. edgetrigger Member

Dec 19, 2010
133
19
your answer is right. Even i am getting the same result. Its almost ten years that i dealt with mesh analysis, but i don't see why i shouldn't concur with you.

3. bglazierjr New Member

Apr 2, 2011
6
1
I know this is a little late but it seems to me that you've mixed up your signs on the second equation

If written using the format approach you will get -I1(-j2)+I2(-j2+j4) = -20j
which simplifies to be I1(j2)+I2(j2) = -20j

Now if you use a calculator such as the ti-89 or ti-86 you can enter the coefficients in directly to solve for the current I1 and I2. A lot of people complain about not being able to do this with an ti-83 or 84 but you actually can with an extremely simple program. If you want I can type the source out.

I1= 3 + 4j -----> 5A< 53.13 degrees
I2= -13-4J -----> 13.6A< -162.9 degrees

I assume you mean the power dissipated in the resistor.
VR1 = 8ohms * 5A< 53.13 degrees = 40 V < 53.13 degrees
40V < 53.13 degrees * 5A< 53.13 degrees = 200 Watts < 106.26 degrees

PV1 = 40V * 5A< 53.13 degrees = 200watts < 53.13 degrees

Last edited: Apr 2, 2011
4. bglazierjr New Member

Apr 2, 2011
6
1
Opps I mean the power is 5/sqrt(2) * 40V / sqrt(2) * cos (0) = 100Watts
and PV1 is 5/sqrt(2) * 40V / sqrt(2) * cos(53.13) = 60 watts

where x= phase shift between voltage and the current for cos(x)