# AC Superposition Question

#### cheyenne

Joined Dec 6, 2007
48
Use superposition to find the voltage VAB?

Attempt:

Firstly I short circuited the voltage source and put an open circuit instead of the current source so there were two circuits to analyse. I thought the best way to do this would be to find the current in AB first in both circuits and then subtract one from the other to get the current. Then use Ohms law to get the voltage AB. I can't seem to get the correct answer.

For the circuit with just the voltage source it is just a straight forward series circuit so I=Vs/Z

I = 5 Angle 90/3 Angle 0 = 1.66 Angle 90 Amps.

Is this correct? Have I made any errors when converting to polar form.

For the circuit with just the current source I thought it would be best to use the current divider rule. I think this is where I'm going wrong.

I = Is(-j4/(3+j4//-j4))

Is this formula correct? If not could someone tell me correct way to do it. I only need help with the currents. Thanks.

#### mik3

Joined Feb 4, 2008
4,846
Use superposition to find the voltage VAB?

Attempt:

Firstly I short circuited the voltage source and put an open circuit instead of the current source so there were two circuits to analyse. I thought the best way to do this would be to find the current in AB first in both circuits and then subtract one from the other to get the current. Then use Ohms law to get the voltage AB. I can't seem to get the correct answer.

For the circuit with just the voltage source it is just a straight forward series circuit so I=Vs/Z

I = 5 Angle 90/3 Angle 0 = 1.66 Angle 90 Amps.

Is this correct? Have I made any errors when converting to polar form.

For the circuit with just the current source I thought it would be best to use the current divider rule. I think this is where I'm going wrong.

I = Is(-j4/(3+j4//-j4))

Is this formula correct? If not could someone tell me correct way to do it. I only need help with the currents. Thanks.
For the circuit with the voltage source why you divided only by the value of the resistor? where are the values of inductor and capacitor impedance? you have to include them also in the calculation

#### cheyenne

Joined Dec 6, 2007
48
I thought the total impedance in the circuit with the voltage source was:

3 Ohms + j4 Ohms + (-j4 Ohms)

Does that become 3 + 0j = 3 Angle 0 in polar form. Is this correct? Thanks

#### mik3

Joined Feb 4, 2008
4,846
I thought the total impedance in the circuit with the voltage source was:

3 Ohms + j4 Ohms + (-j4 Ohms)

Does that become 3 + 0j = 3 Angle 0 in polar form. Is this correct? Thanks
you are correct yes

#### cheyenne

Joined Dec 6, 2007
48
Can you explain to me how current division would work in this example? I think that is where I am going wrong.

#### mik3

Joined Feb 4, 2008
4,846
Can you explain to me how current division would work in this example? I think that is where I am going wrong.
lets say you want to find the current through the capacitor

Ic=Io(current source)*(impedance of the inductor+resistance of the resistor)/
(impedance of the inductor+impedance of the capacitor+resistance of the resistor)

ok?