Answer Posted / dhanraj

For 33 Kv cable DC hi pot test can be done like (2*E+1)1.414
here E=Voltage , 1.414 is the root 2 value.
Now (2*33+1)=67Kv
Then 67*1.414=94.738 Kv this is 100% of cable test.
In manufacturing factory they will test up to 80% of the voltage in that 100% (94.73*0.8)=75.79 Kv.
But in consumer side they will test up to 80% of voltage that injected in the factory voltage (75.79*0.8)=60.64 Kv.
Now 60.64 Kv is apply for 3 mins maximum.

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