A `100 mu F` capacitor in series with a `40 Omega` resistance is connected to a 110V. 60 Hz supply.
What is the time lag between the current maximum and the voltage maximum ?

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1 Answers

Given capacitance of capacitor
`C = 100 mu F=100xx10^(-6)F`
Resistance `R = 40 Omega`
the rms value of voltage `V_("rms")=110 V`
Frequency f = 60 Hz
Time lag `(t) = (phi)/(omega)`
`tan phi =(X_(C ))/(R )=(1)/(2pi fCR)`
`= (1)/(2xx3.4xx60xx10^(-4)xx40)`
`phi=tan^(-1)(0.6628)=33.5^(@)=(33.5pi)/(180)`
Time lag `(33.5pi)/(180xx2pixx60)=1.55xx10^(-3)S`
Thus, the time lag between the voltage maximum and the current maximum is `1.55xx10^(-3)S`.

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