A `100muF` capacitor in series with a `40 Omega` resistor is connected to a `110V`, `60 HZ` supply. What is the time lag between current maximum and voltage maximum?

Explanation

Capacitance of the capacitor, ` C=1 00muF=100xx10^(-6)F`
Resistance of the ressitor, R=`40Omega`
Supply voltage , `V=110V`
(a) Frequency of oscillations, v=60Hz
Angular frequency, `omega=2piv=2pixx60` rad/s
For a RC circuit, we have the relation for impedance as:
`Z=R^(2)+(1)/(omega^(2)C^(2))`
peak voltage, `v_(0)=vsqrt(2)=110sqrt(2)V`
Maximum current is given as:
`I_(@)=(V_(@))/(Z)`
`=(V_(@))/(sqrt(R^(2)+(1)/(omega^(2)C^(2)))` ltbgt `=(110sqrt(2))/(sqrt((40^(2)+(1)/((120pi)^(2)xx(10^(-4))^(2)))`
`=(110sqrt(2))/(sqrt(1600+(10^(8))/((120pi)^(2)))=3.24A`
(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of `phi`. this angle is given by the relation.
`therefore phi=(omegaC)/(R)=(1)/(omegaCR)`
`(1)/(120pixx10^(-4)xx40)=0.6635`
`phi=tan^(-1)(0.6635)=33.56^(@)`
`=(33.56pi)/(180)` rad
`therefore `time lag `=(phi)/(omega)`
`=( 33.56 pi)/(180xx 120pi)=1.55xx10^(-3)s=1.55ms`
Hence, the time lag between maximum current and maximum voltage is 1.55ms.