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

Explanation

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
Impedance Z
`= sqrt(R^(2)+X_(C )^(2))=sqrt(R^(2)+((1)/(2pi fC))^(2))`
`= sqrt((40)^(2)+((1)/(2xx3.14xx60xx10^(-6)xx100)))`
`= sqrt(1600+704.33) = 48 Omega`
The rms value of current
`I_("rms")=(V_("rms"))/(Z)=(110)/(48)`
Maximum current in the circuit
`I_(0)= sqrt(2)I_("rms")=1.414xx(110)/(48)=3.24A`