A 60 Ω resistor, a 1·0 H inductor and a 4μF capacitor are connected in series to an ac supply…

A $60\,\Omega$ resistor, a $1\cdot0\text{ H}$ inductor and a $4\mu\text{F}$ capacitor are connected in series to an ac supply generating an emf $e = 300\sin(500t)\text{ V}$. Calculate:

(a)[2]

impedance of the circuit.

(b)[1.5]

peak value of the current flowing through the circuit.

(c)[1.5]

phase difference between the current and the supply voltage.

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Answer

Answer (a)

Formula used: Inductive reactance: XL = 2*pi*f*L with L = 1 H, f = 79.58 Hz, so XL = 500 ohm

$\omega = 500\text{ rad s}^{-1}$ $X_L = \omega L = 500\times1.0 = 500\ \Omega$ $X_C = \dfrac{1}{\omega C} = \dfrac{1}{500\times4\times10^{-6}} = 500\ \Omega$ $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{60^2 + 0} = 60\ \Omega$ (the circuit is in resonance)

Final answer: 60 ohm

Answer (b)

Peak emf $e_0 = 300\text{ V}$. $i_0 = \dfrac{e_0}{Z} = \dfrac{300}{60} = 5\text{ A}$

Final answer: 5 A

Answer (c)

Formula used: Phase angle of a series LCR circuit (current lags by phi): phi = atan((XL - XC)/R) with R = 60 ohm, XC = 500 ohm, XL = 500 ohm, so phi = 0 deg

$\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{500 - 500}{60} = 0$, so $\phi = 0^\circ$

Final answer: 0 degree