Question

A `100 mu F` capacitor in series with a `40 Omega` resistor is connected to a 100 V, 60 Hz supply. The maximum current in the circuit is
A. `2.65 A`
B. `2.75 A`
C. `2.85 A`
D. `2.95 A`

Answer 1

Correct Answer - D
Here, `C=100 mu F=100xx10^(-6)F=10^(-4)F`,
`R=40 Omega, V_("rms")=100 V, upsilon = 60 Hz`
`therefore V_(0) = sqrt(2)V_("rms")=100 sqrt(2)V`
In series RC circuit, `Z=sqrt(R^(2)+X_(C )^(2))=sqrt(R^(2)+(1)/(omega^(2)C^(2))) " " [because omega = 2pi upsilon)`
Maximum current in the circuit,
`I_(0)=(V_(0))/(Z)=(V_(0))/(sqrt(R^(2)+(1)/(4pi^(2)upsilon^(2)C^(2))))`
`=(100sqrt(2))/(sqrt((40)^(2)+(1)/(4xx(3.14)^(2)xx(60)^(2)xx(10^(-4))^(2))))`
`=(100sqrt(2))/(48)=2.95 A`