Solve Superposition Theorem Homework: 50Ω Load

[gneill]

Thank you for the guidance gneill I must have finger bashed my calculator on the first part. All looks as it should now

Excellent

 

[Student12345]

Hi I was hoping someone can help me with part c)
I seem to be getting negative current and I'm not sure if this is correct
So far I have
I1= v1/j4
415/j4
(415)(-j4)/(j4)(-j4)
-j1660/16
I1= -j103.75

I2=v2/j6
-j415/j6
(-j415)(-j6)/(j6)(-j6)
-2490/36
I2=-69.167

Not sure where to go after this can I add the 2 currents together and add the 2 parallel reactances (1/x1+1/x2=1/xtotal)?

 

[gneill]

Not sure where to go after this can I add the 2 currents together and add the 2 parallel reactances (1/x1+1/x2=1/xtotal)?

Sure.

 

[Student12345]

Sure.

I have got an answer but it doesn't match my answers from a) and b) can you please guide me in the direction of where I am going wrong?

My calculations give I1 =-j103.75 and I2=-69.167 as shown in the post above. I then formed Zn=-j2.4 which I believe is correct.

I then used IL= In * Zn/Zn+Zl and I think this is where I went wrong.
IL= (-69.167-j103.75)* (-j2.4/(-j2.4*(35+J35.71)))

IL= (-69.167-j103.75)*(-j2.4/(85.704-j84)

IL= (-69.167-j103.75)*(0.01399-j0.01428)

IL=-0.51-j0.46

Obviously I have calculations inbetween.

 

[gneill]

I have got an answer but it doesn't match my answers from a) and b) can you please guide me in the direction of where I am going wrong?

My calculations give I1 =-j103.75 and I2=-69.167 as shown in the post above. I then formed Zn=-j2.4 which I believe is correct.

The currents are good but the sign on your combined impedance is not.

I then used IL= In * Zn/Zn+Zl and I think this is where I went wrong.
IL= (-69.167-j103.75)* (-j2.4/(-j2.4*(35+J35.71)))

IL= (-69.167-j103.75)*(-j2.4/(85.704-j84)

You have multiplied the impedance terms in the denominator rather than adding them!

 

[Student12345]

The currents are good but the sign on your combined impedance is not.

You have multiplied the impedance terms in the denominator rather than adding them!

Haha so I did. Thanks for that!

Zn I got from
1/Zn = 1/z1+1/z2 = 1/j4+1/j6
=(-j4/(j4)(-j4))+(-j6/(j6)(-j6))
=(-j4/-j²16)+(-j6/-j²36)
=(-j4/16)+(-j6/36)
1/Zn=-j0.417 Zn=-j2.4
Where have I gone wrong?

 

[gneill]

Haha so I did. Thanks for that!

Zn I got from
1/Zn = 1/z1+1/z2 = 1/j4+1/j6
=(-j4/(j4)(-j4))+(-j6/(j6)(-j6))
=(-j4/-j²16)+(-j6/-j²36)
=(-j4/16)+(-j6/36)
1/Zn=-j0.417 Zn=-j2.4
Where have I gone wrong?

Your last step where you take the reciprocal does not handle the j properly.

 

[Student12345]

Your last step where you take the reciprocal does not handle the j properly.

Ok thanks I'm a little confused because normally if I have a figure say -j²35.71² I will change the sign and drop the j² to give +35.71²but it doesn't seem to be right for this one

 

[gneill]

$Zn = \frac{1}{-j 0.417} = \frac{2.4}{-j} \cdot \frac{j}{j} = \frac{j 2.4}{+1} = j 2.4$

The bottom line being that when you "move" a j from denominator to numerator (or vice versa) you change its sign

 

[Student12345]

$Zn = \frac{1}{-j 0.417} = \frac{2.4}{-j} \cdot \frac{j}{j} = \frac{j 2.4}{+1} = j 2.4$

The bottom line being that when you "move" a j from denominator to numerator (or vice versa) you change its sign

Thanks so much I think I've been staring at this question so long it's driven me crazy!