Question

A series CR circuit with R = 200 Ω and C = (50/π) µF is connected across an ac source of peak voltage ε₀ = 100 V and frequency v = 50 Hz. Calculate (a) impedance of the circuit (Z), (b) phase angle (Φ), and (c) voltage across the resistor.

Answer 1

(a) Given: R = 200Ω, C = `50/pi` μF, ε₀ = 100V, Frequency(ν) = 50Hz

We know that the reactance of a capacitor (X_C) is given by the formula.

X_C = `1/((omegaC))`

where C is the capacitance and ω = 2πν.

∴ X_C = `1/(omegaC) = 1/(2piνC)`

X_C =  `1/((2pi xx 50 xx 50"/"pi xx 10^-6))`

X_C = `10^6/((2 xx 50 xx 50))`

X_C = 200 Ω

R = 200 Ω

The impedance of the circuit can be calculated by the given formula,

`Z^2 = R^2 + X_{C^2}`

`Z^2 = (200)^2 + (200)^2`

Z = `200sqrt2Omega`

(b) The phase angle can be calculated as

`tanphi = X_C/R`

`tanphi = 200/200`

`tanphi` = 1

Φ = 45°

(c) Voltage across the resistor can be calculated as,

ε₀ = 100 V

`I_("r.m.s.") = E_("r.m.s.")/Z = 100/(200sqrt2)`

`(V_"r.m.s.")_R = I_("r.m.s.") xx R`

`(V_"r.m.s.")_R = (E_"r.m.s." xx R)/Z`

`(V_"r.m.s.")_R = 100/((sqrt2 xx 200sqrt2)) xx 200`

`(V_"r.m.s.")_R = 100/2`

`(V_"r.m.s.")_R = 50` V