Question

A 60 Ω resistor, a 1.0 H inductor and a 4 μF capacitor are connected in series to an ac supply generating an emf e = 300 sin (500t) V. Calculate:

1. impedance of the circuit.
2. peak value of the current flowing through the circuit.
3. phase difference between the current and the supply voltage.

Answer 1

(a) Given that C = 4 μF = 4 × 10^-6 F

R = 60 Ω; L = 1.0 H

and e = 300 sin (500t) V

Since e = e₀ sin ωt

Then, by comparing

e₀ = 300 V, ω = 500 Hz

∴ XL = ωL = 500 Ω

and `"X"_"C" = 1/(omega"C") = 1/(2pi"fc")`

`= 1/(500 xx 4 xx 10^-6)`

= 500 Ω

∴ Z = `sqrt(("X"_"L" - "X"_"C")^2 + "R"^2) = sqrt("R"^2)`

R ≈ 60.00 ohm.

(b) Peak value of current

`"i"_0 = "e"_0/"Z"`

`= 300/60`

= 8.5 A

(c) Phase difference `phi = cos^-1  "R"/"Z"`

`= cos^-1  60/60`

= `0^circ`