# A single phase 10KVA 50Hz transformer 1 KV primary winding with current 0.5A and 55W at the rated voltage and frequency, on no load. A second transf

147 views
in General
closed

A  single phase 10KVA 50Hz transformer 1 KV primary winding with current 0.5A and 55W  at the rated voltage and frequency, on no load. A second transformer has a core with all its linear dimensions √2 times the corresponding dimensions of the first transformer. The core thickness and laminations are the same in both the transformers. The primary winding of both the transformers have same number of turns. If the rated voltage of 2KV at 50Hz is applied to the primary of the second transformer, the no load current and power are:

1.

0.7A,77.8 W

2.

0.7A,155.6 W

3.

1 A,110 W

4.

1 A,220 W

by (37.3k points)
selected

Correct Answer - Option 2 :

0.7A,155.6 W

Core loss α core volume

∴ New loss ${\left( {\sqrt 2 } \right)^3} \times 55 = 155.6W$

Core loss ${P_i} = V{I_c}$

$\begin{array}{l} {I_{{c_1}}} = \frac{{55}}{{1000}} = 0.055\ A\\ {I_{{c_2}}} = {\left( {\sqrt 2 } \right)^3}{I_{{c_1}}} \end{array}$

From the data for 2nd transformer

$\begin{array}{l} {I_{{c_2}}} = 0.155\ A\\ \therefore {I_{{m_1}}} = 0.496\left[ {\therefore \sqrt {{{\left( {0.5} \right)}^2} - {{\left( {0.055} \right)}^2}} = 0.496} \right] \end{array}$

Reluctance ${R_2} = \frac{{{R_1}}}{{\sqrt 2 }}$

${\emptyset _{{m_1}}} = \frac{{4000}}{{4.44{N_1}f}}$             ${\emptyset _{{m_2}}} = \frac{{2000}}{{4.44\ {N_1}f}}$

$\begin{array}{l} {\emptyset _{{m_2}}} = 2{\emptyset _{{m_1}}}\\ {\emptyset _{{m_1}}} = \frac{{{N_1}{I_m}_2}}{{{R_1}}}\\ {\emptyset _{{m_2}}} = \frac{{{N_1}{I_m}_2}}{{\frac{{{R_1}}}{{\surd 2}}}}\\ \therefore {I_{{m_2}}} = \surd 2{I_{{m_1}}}\\ = \sqrt 2 \times 0.496 \end{array}$

= 0.702 A

$\begin{array}{l} \therefore {I_0} = \sqrt {I_{{m_1}}^2 + I_{{c_2}}^2} \\ = \sqrt {{{\left( {0.7025} \right)}^2} - {{\left( {0.155} \right)}^2}} = 0.718\ A \end{array}$