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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:
- impedance of the circuit.
- peak value of the current flowing through the circuit.
- phase difference between the current and the supply voltage.
Solution
(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`
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