Superposition on circuits with impedance elements

[Chickwolf]

Hey guys, running through an example on superposition but have struck a little hiccup. Have attached the sheet of workings as given out by my lecturer, I understand it up to the i2a calculation, but I cant get my head around why i1a is being multiplied by what seems to be half the equation for the parallel resistors. Would this not give a voltage value? Have been scratching my head at this for the past 20 minutes so thought I would check in and see if it just a stupid mistake I have made or am just missing something.

Cheers for the help guys!

 

[Ratch]

Hey guys, running through an example on superposition but have struck a little hiccup. Have attached the sheet of workings as given out by my lecturer, I understand it up to the i2a calculation, but I cant get my head around why i1a is being multiplied by what seems to be half the equation for the parallel resistors. Would this not give a voltage value? Have been scratching my head at this for the past 20 minutes so thought I would check in and see if it just a stupid mistake I have made or am just missing something.

Cheers for the help guys!

I see I1a being multiplied by a ratio according to the Current Division Law. It calculates the current present in the rightmost branch. Why do you show the red current arrow direction in the rightmost branch opposing the direction of the current exiting 10 volt source?

Ratch

 

[Chickwolf]

I see I1a being multiplied by a ratio according to the Current Division Law. It calculates the current present in the rightmost branch. Why do you show the red current arrow direction in the rightmost branch opposing the direction of the current exiting 10 volt source?

Ratch

The red arrows were attached by the lecturer when he released the notes, simple muck up from him I imagine. As we have the source current and voltage, could we not simply just work out the voltage drop over the 1 ohm resistor, then use that new voltage to determine the current over the 2 ohms resistor?

Edit: Was just working through that part using the current divider formula but his workings doesn't seem to match up with what I had in my notes. It was my understanding that the current formula was In = Itotal * (Rtotal / Rn) , but he seems to have something a bit different? I thought we would work out the total current, then the current over the 2 ohm resistor, and use the two of these to work out the current over the center impedance? Am sorry if this is quite an obvious thing, has been a while since I have done current dividers!

 

[Ratch]

The red arrows were attached by the lecturer when he released the notes, simple muck up from him I imagine. As we have the source current and voltage, could we not simply just work out the voltage drop over the 1 ohm resistor, then use that new voltage to determine the current over the 2 ohms resistor?

Edit: Was just working through that part using the current divider formula but his workings doesn't seem to match up with what I had in my notes. It was my understanding that the current formula was In = Itotal * (Rtotal / Rn) , but he seems to have something a bit different? I thought we would work out the total current, then the current over the 2 ohm resistor, and use the two of these to work out the current over the center impedance? Am sorry if this is quite an obvious thing, has been a while since I have done current dividers!

https://www.google.com/search?q=cur...3.69i57j0l5.5108j0j8&sourceid=chrome&ie=UTF-8

Ratch

 

[Chickwolf]

Why is it that in his sum, he has put Rn on the top, and then the sum of the other two resistors on the bottom? Is that not just the opposite of the formula?

 

[Ratch]

Why is it that in his sum, he has put Rn on the top, and then the sum of the other two resistors on the bottom? Is that not just the opposite of the formula?

That question is unnecessary to the problem solution. You have the link which shows the correct way to apply the current division method. You should question the person who gave you the wrong information.

Ratch

Why is it that in his sum, he has put Rn on the top, and then the sum of the other two resistors on the bottom? Is that not just the opposite of the formula?

In a parallel circuit, the total resistance is less that the resistance of any branch. If Rn were on the bottom, the calculation for the current in a branch would be greater in value than the supply current. That would be impossible. Study the material in the link I sent you.

Ratch

 

[Laplace]

Rather than memorize a formula for current division, one could employ a method where the general format for voltage division and for current division is identical. Voltage division is proportional to resistance {impedance} Ω, whereas current division is proportional to conductance {admittance} ℧. A=1/Z. In the following figure note the similarity of the equations for voltage division in terms of Z and for current division in terms of A.

While the problem statement required solution using superposition and current division, when does such clever methodology present any advantage over conventional nodal analysis? For comparison purposes, this is the solved node equation for the circuit, using the node voltage and branch impedance to calculate the current in each branch, as shown.

Here is the same problem analyzed with superposition and current division. Now contrast this rat's nest of complex arithmetic with the previous nodal analysis.