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A `50 Hz` alternating current of peak value `1` ampere flows through the primary coil of a transformer. If the mutual inductance between the primary secondary be `1.5` henry, then the peak value of the induced voltage isA. `75 V`B. `150 V`C. `225 V`D. `300 V` |
Answer» Correct Answer - D `T=1/50 s` For the current to rise from zero to peak value, required time is `T/4` i.e. `1/200 s`. Now `E=1.5 (1-0)/(1//200) V=1.5xx200V=300 V`. |
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