Power Loss in transmission lines: Which method is correct?

[PaulK]

Here is the problem:
http://postimage.org/image/1jy6nol5w/

Method 1:

Steps
1. Calculate V(T1)=12*10=120V,
V(T2)=10*10=100V
2. Since V(3)=10V and I(3)=1A, P(T3)=10W
3. Since not otherwise stated, Assume transformers are perfect.
4. So P(T3)= P(T2) = 10W
5. known P(T2)=10w and V(T2) =100V, Calculate: I(T2)= 10/100 = 0.1 A
6. Since resistors are connected in series, Current through each resistor = 0.1A
7. Power Loss = (0.1)^2 * R = 0.001 * 16 = 0.16WMethod 2.
1. Calculate V(T1)=12*10=120V,
V(T2)=10*10=100V
2. Voltage consume by resistors = V(T1)- V(T2) = 120 -100 =20V
3. Power loss = (20)^2 /16 = 25W
Both method seem to make sense, but of course only one of the is right (Or none?)
Please tell me which one is wrong and why.

Thanks a lot.

 

[berkeman]

Here is the problem:
http://postimage.org/image/1jy6nol5w/

Method 1:

Steps
1. Calculate V(T1)=12*10=120V,
V(T2)=10*10=100V
2. Since V(3)=10V and I(3)=1A, P(T3)=10W
3. Since not otherwise stated, Assume transformers are perfect.
4. So P(T3)= P(T2) = 10W
5. known P(T2)=10w and V(T2) =100V, Calculate: I(T2)= 10/100 = 0.1 A
6. Since resistors are connected in series, Current through each resistor = 0.1A
7. Power Loss = (0.1)^2 * R = 0.001 * 16 = 0.16W


Method 2.
1. Calculate V(T1)=12*10=120V,
V(T2)=10*10=100V
2. Voltage consume by resistors = V(T1)- V(T2) = 120 -100 =20V
3. Power loss = (20)^2 /16 = 25W



Both method seem to make sense, but of course only one of the is right (Or none?)
Please tell me which one is wrong and why.

Thanks a lot.

I think there is an error in the problem statement. They are over-constraining the problem by giving you an output voltage of 10V.

Start with the input voltage and output current (like in your First Method). The output current defines the TL current of 0.1A, which defines the voltage drops across the TL resistors. The input voltage and the TL voltage drops define the transformed output voltage. What do you get for the transformed output voltage? Not 10V.

 

[PaulK]

It is such a relief. Thanks a lot, berkeman.