An a.c. source of 200 V, 50 Hz is connected across a `400 Omega` resistor and capacitor of 25pF in series. Calculate reactance

Explanation

`E_v = 200 V`
Frequency v = 50 Hz
Resistance, `R = 300Omega`
Capacitance, `C = 25/pimuF = 25/pixx10^(-6)F`
Resistance `(X_c) = ?`
Impedance (Z) = ? and current `(I_v) = ? `
(i) Reactance = `X_c`
`=1/(2pivC)=1/(2pixx50xx25/pixx10^(-6))`
`X_C = 10^6/(2500)=400Omega`
(ii) Impedance = Z
`=sqrt(R^2+X_c^2)= sqrt((300)^2+(400)^(2))`
`Z = sqrt(90000+160000) = sqrt(250000)`
`Z = 500Omega`
(iii) `I=E_v/Z = 200/500=0.4A`