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An ideal choke draws a current of 8A when connected to an AC supply of 100 V, 50 Hz. A pureresistor draws a current of 10 A when connected to the same source. The ideal choke and the resistorare connected in series and then connected to the AC source of 150 V, 40 Hz. The current in thecircuit becomes : R=(100)/(10)=10 Omega X_(L)=(100)/(8)=12.5=2pi fL |
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Answer» `=(15)/(sqrt(2))A` `X_(L)=(100)/(8)=12.5=2pi FL` `L=0.04 H` `X'_(L)=2pi f'L = 10 Omega` `Z'=sqrt(R^(2)+X'_(L)^(2))=10 sqrt(2)Omega` `I=(150)/(10sqrt(2))=(15)/(sqrt(2))A` |
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