Engineer Boards

​ ​ ​

# Complex Imaginary Volume 1, Problem 22

## Recommended Posts

This problem gives the following information regarding two parallel transformers: Impedance of T1 = 5.25%, Impedance of T2 = 4.75%, total current =11,500A.

The question then asks for T1's current contribution.

I get:

I = 11500*(0.0525/(0.0525+0.0475) = 6037.5A

However, the answer provided uses the impedance of T2 in the numerator, and gives 5462.5A.

Why would the impedance of T2 be used to calculate T1's current contribution?

##### Share on other sites

I dunno.  That's unreal.

• 1

##### Share on other sites

Glad you had the same reaction as me - I'm assuming its an error in the solutions. Just needed a sanity check.

##### Share on other sites
On 9/2/2018 at 9:02 AM, ellen3720 said:

This problem gives the following information regarding two parallel transformers: Impedance of T1 = 5.25%, Impedance of T2 = 4.75%, total current =11,500A.

The question then asks for T1's current contribution.

I get:

I = 11500*(0.0525/(0.0525+0.0475)﻿ = 6037.5A

However, the answer provided uses the impedance of T2 in the numerator, and gives 5462.5A.

Why would the impedance of T2 be used to calculate T1's current contribution?

1

Hi @ellen3720,

When two transformers are connected in parallel, the problem is very similar to a standard current division circuit from circuit analysis. I find that it helps to see why Z2 ends up in the numerator to really understand this problem.  Let's derive it.

In the example you posted above, the total load current will equal the sum of T1's and T2's current contribution since they are in parallel (KCL relationship):

• I_L = I_1 + I_2

Assuming both transformers have the same ratio and voltage (V1 = V2 = V), we can substitute Ohm's law for I_1 and I_2 using the impedance of each transformer (Z1 and Z2) and the voltage (V):

• I_L = I_1 + I_2
• I_L = V/Z1 + V/Z2

Next, we can add both fractions:

• I_L = V/Z1 + V/Z2
• I_L = (V/Z1)•(Z2/Z2) + (V/Z2)•(Z1/Z1)
• I_L = (V•Z2)/(Z1•Z2) + (V•Z1)/(Z1•Z2)
• I_L = V(Z1+Z2) / (Z1•Z2)

Then, plug in I_1 = V/Z1 since we are solving for I_1:

• I_L = V(Z1+Z2) / (Z1•Z2)
• I_L = I_1 • (Z1+Z2) / Z2

Now, solve for I_1 (This is where Z2 ends up in the numerator):

• I_1 = I_L •  Z2/(Z1+Z2)

Finally, we can plug in the values from the problem and check our answer with your provided solution:

• I_1 = I_L •  Z2/(Z1+Z2)
• I_1 = 11,500A •  4.75%/(5.25%+4.75%)
• I_1 = 5462.5A

Hope this helps.

##### Share on other sites

Zach,

Thanks so much  - that makes sense, and I never would have figured it out on my own!

##### Share on other sites

its been less than 2 years since i passed that exam and i don't remember shit 😫