Auto-transformer makes effective saving on copper and copper losses, when its transformation ratio is

Copper saving in autotransformer

Consider a 2-winding transformer reconnected as an auto-transformer as shown below:

The turns ratio of an auto-transformer is given by:

\(a={N_1\over N_2}={V_1\over V_2}\)

where, N₁ = No. of turns of high voltage side

N₂ = No. of turns of low voltage side

The copper used in a transformer is directly proportional to the MMF.

For an ideal auto-transformer: \(N_1I_1=N_2I_2\)

The MMF of an auto-transformer is given by:

\(MMF=(N_1-N_2)I_1\space +\space N_2(I_2-I_1)\)

\((MMF)_{auto}=N_1I_1-N_2I_1+N_2I_2-N_2I_1\)

\((MMF)_{auto}=2(N_1I_1-N_2I_1)\).........(i)

The MMF of a 2-winding transformer is given by:

\((MMF)_{2-wdg}=N_1I_1+N_2I_2\)

\((MMF)_{2-wdg}=2N_1I_1\) ........(ii)  (By MMF balance equation- \(N_1I_1=N_2I_2\))

Dividing equation (i) by (ii), we get:

 \((MMF)_{auto}=({1-{N_2\over N_1}})\times(MMF)_{2-wdg}\)

The percentage saving of copper used in auto-transformer is:

% saving = \({(MMF)_{2-wdg}-(MMF)_{auto}\over (MMF)_{auto}}\)

% saving = \({N_2\over N_1}\)

Since the value \(a={N_1\over N_2}={V_1\over V_2}\) is slightly greater than 1.

Therefore, the value of % saving = \({N_2\over N_1}={1\over a}\) is approximately equal to one.