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In a series LR circuit, the voltage drop across inductor is 8 volt and across resistor is 6 volt. Then voltage applied and power factor of circuit respectively are:A. Volatage of the source will be leading current in the ciruitB. Volatage drop across each element will be less the appliced voltageC. Power factor of circuit will be 4/3D. None of these |
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Answer» Correct Answer - D Since cos `theta = (R)/(Z) = (IR)/(IZ) = (8)/(10) = (4)/(5) ` (cos `theta` can never be greater than 1) Also , `IX_(C) gt IX_(L) implies X_(C) gt X_(L)` Current will be leading In a LCR circuit `V = sqrt((V_(L) - V_(C))^(2) + V_(R)^(2)) = sqrt((6 - 12)^(2) + 8^(2))` V = 10 , which is less than voltage drop across capacitor . |
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