Wednesday, June 10, 2009

Electrical Interview questions Part 9

Q:How to calculate capacitor bank value to maintain unity power factor with some suitable example?
A:KVAR= KW(TAN(COS(-1)#e)- TAN(COS(-1)#d) )

Q:Tell me in detail about c.t. and p.t. ?(Company:reliance)
A:The term C.T means current transformer,and the term P.T means potential transformer.In circuit where measurements of high voltage and high current is involved they are used there.Particularly when a measuring device like voltmeter or ammeter is not able to measure such high value of quantity because of large value of torque due to such high value it can damage the measuring, CT and PT are introduced in the circuits. They work on the same principle of transformer, which is based on linkage of electromagneticflux produced by primary with secondary.They work on the ratio to they are designed.E.g if CTis of ratio 5000\5A and it has to measure secondary current of
8000A.then ANS=8000*5\5000=8Aand this result will be given to ammeter .and after measuring 8A we can calculate the primary current.same is
the operation of PT but measuring voltage.

Q:There are a Transformer and an induction machine. Those two have the same supply. For which device the load current will be maximum? And why?
A:The motor has max load current compare to that of transformer because the motor consumes real power.. and the transformer is only producing the working flux and its not consuming.. hence the load current in the transformer is because of core loss so it is minimum.

Q:what is power factor? whether it should be high or low? why?
A:Power factor should be high in order to get smooth operation of the system.Low power factor means losses will be is the ratio of true power to apperent power. it has to be ideally 1. if it is too low then cable over heating & equipment overloading will occur. if it is greater than 1 then load will act as capacitor and starts feeding the source and will cause tripping.(if pf is poor ex: 0.17 to meet actual power load has to draw more current(V constant),result in more lossesif pf is good ex: 0.95 to meet actual power load has to draw less current(V constant),result in less losses).