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A `0.2 k Omega` resistor and `15 mu F` capacitor are connected in series to a 220 V, 50 Hz ac source. The impadance of the circuit isA. `250 Omega`B. `268 Omega`C. `29.15 Omega`D. `291.5 Omega` |
Answer» Correct Answer - D Here, `R=0.2 k Omega = 200 Omega` `C =15 mu F = 15xx10^(-6)F, V_("rms")=220 V, upsilon = 50 Hz` Capacitive reactance, `X_(C )=(1)/(2pi upsilon C)=(1)/(2xx3.14xx50xx15xx10^(-6))=212 Omega` The impedance of the Rccircuit is `Z = sqrt(R^(2)+X_(C )^(2))=sqrt((200)^(2)+(212)^(2))=291.5 Omega` |
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