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A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a `220` V source. When the capacitor is `50%` charged, the peak value of the displacement current isA. 2.2 AB. 0.45833333333333C. 4.4 AD. `11sqrt(2)` A |
Answer» Correct Answer - A Impedance of the R-C circuit, `Z= sqrt(R^(2)+X_(C)^(2))` Where, `R=100Omega` and `X_(C) = 100Omega` `rArr Z=sqrt((100)^(2)+(100)^(2))` = `100sqrt(2)Omega` Peak value of the current, `I_("max") = V_("max")/Z = (220sqrt(2))/(100sqrt(2))`=2.2 A |
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