A resistor of 200 `Omega` and a capacitor of 15.0`muF` are connected in series to a 22V. 50Hz ac source. (a) Calculate the current in the circuit. (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? if yes, resolve the paradox.

Explanation

Given, `R=200 Omega, C=15.0(mu)F=15.0xx10^(-6)F`
`E_(rms)=220V,f=50Hz`
(a) In order to calculate the current, we need the impenance of the circuit.
`Z=sqrt(R^(2)+X_(L)^(2))=sqrt(R^(2)+(2 pi g C)^(-2))`
`=sqrt((200 Omega)^(2)+(2xx3.14xx50xx15xx10^(-6)F)^(-2))`
`sqrt((200 Omega)^(2)+(212 Omega)^(2))=291.5 Omega`
Therefore, the current in the circuit is
`I_(rms)=(E_(rms))/(Z)=(220V)/(291.5 Omega)=0.755A`.
(b) Since the current is hte same throughout the circuit.
`V_(R )=I_(rms)R=(0.755A)(200 Omega)=151 V`
`V_(C )=I_(rms)X_(C )=(0.755A)(212 Omega)=160 V`.
The algebraic sum of hte two voltages `V_(R )and V_(C ) is 311 V` which is more than the source voltage of 220V. How to resolve this parodox? You may recall taht the tawo voltages are not in the same phases. Therefore, they cannot be added like the ordinary numbers. The two voltages are out of phase by `90^(@)`. Therefore, the total of these voltages must be obtained using the Pythagoras theoram:
`V_(R+C)=sqrt(V_(R )^(2) + V_(C )^(2))=sqrt((151V)^(2)+(160V)^(2))=220V`
Thus, if the phase difference between two voltages is properly taken into account the total voltage across the resistor and the inductor is equal to the voltages of the source.