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A 50KVA, 2300/230V, 50Hz transformer has a high voltage winding resistance of 0.65ohm and low voltage resistance of 0.0065ohm and this transformer has been tested to determine, its equivalent circuit, the laboratory tests showed the following results.

Open circuit test:

V(oc):2300v 
I(oc):5.7A 
P(oc):190W 

Short circuit test:

V(sc):41.5v 
I(sc):21.7A 
P(sc):Not measured 

*All the data were taken from the high voltage side of the transformer.

Find the equivalent circuit of the transformer referred to the primary side.

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    \$\begingroup\$ This seems to be a homework question. Please post your attempt to solve the problem, along with any insight you have as to how you think you might solve the problem. \$\endgroup\$
    – uint128_t
    Commented Jan 13, 2016 at 6:38
  • \$\begingroup\$ I'm voting to close this question as off-topic because its homework \$\endgroup\$
    – brhans
    Commented Jan 18, 2016 at 1:32

1 Answer 1

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The Power which is utilized by the power transformer during short circuit test is approximately equal to copper loss of transformer. Which is mainly due to resistance of copper winding.

During short circuit test Iron loss are negligible compare to copper loss.

Approximate Reading in the meter might be (which is equal to copper loss)

$$= I(sc)2 \times R_{total}$$

where

\$R_{total} = \$ Total winding resistance of transformer referred to primary.

\$R_{total} = \$ Primary resistance + {secondary resistance \$ \times \Big(\dfrac{V_1}{V_2}\Big)^2\$ }

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