NEETPhysicsAlternating CurrentMCQ+4 / −1
A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drwn by the primary windings of the transformer is
- A3.6 ampere
- B2.8 ampere
- C2.5 ampere
- D5.0 ampere
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Correct answer: D
Efficiency of the transformer,
=
Output power
Input power
$ \Rightarrow $ $${{80} \over {100}} = {{{V_s}{I_s}} \over {{V_p}{I_p}}}$$
$ \Rightarrow $ Ip = $${{440 \times 2} \over {{{80} \over {100}} \times 220}}$$ = 5 A
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