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A resistor of 200 Omega and a capacitor of 15.0 muF are connected in series to a 220V, 50Hz as source . (a) Calculate the current in the circuit. (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the alebraic sum of these voltages more than the source voltage? If yes resolve the paradox. |
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Answer» SOLUTION :Here, for series RC a.c. circuit, `I_(rms) = ( V_(rms))/(sqrt(R^(2) + X_(C )^(2))) = ( V_(rms))/( sqrt(R^(2) + ( 1)/( 4 pi^(2) f^(2) C^(2))))` `:. I_(rms)= ( 220)/( sqrt((200)^(2) + (1)/( 4 xx (3.14)^(2) xx ( 50)^(2) xx ( 15 xx 10^(-6))^(2))))` `:. I_(rms) = 0.755A` (b) (i) `V_(R )- I_(rms) R ` `= ( 0.77) ( 200)` = 151 Volt (ii) `V_(C ) = I_(rms) X_(C )` `= I_(rms)xx (1)/( 2PI f C )` `= ( 0.755) xx (1)/( (2) ( 3.14) ( 50) ( 15 xx 10^(-6))) ` = 160.3 Volt (c ) Here `V_(R )` an `V_(C )` are not in same PHASE. Hence we can not MAKE their algebraic sum to get resultant voltage. Actually, phase difference between` V_(R )` and `V_(C )` is `( pi )/(2)` rad and so resultant voltage can be found out from following phasor diagram. `V = sqrt( V_(R )^(2) + V_(C )^(2))` `= sqrt((151)^(21) + ( 160.3)^(2))` `:. V = 220 `Volt  Above voltage is equal to rms value of source voltage. |
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