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Answer the following.
A 220 V, 50 Hz ac source is connected to a coil having coefficient of self-inductance of 1 H and a…
(a)[2.5]
A $220\text{ V}$, $50\text{ Hz}$ ac source is connected to a coil having coefficient of self-inductance of $1\text{ H}$ and a resistance of $400\,\Omega$. Calculate:
(1) the reactance of the coil.
(2) the impedance of the coil.
(3) the current flowing through the coil.
(b)[2.5]
Draw a labelled graph showing variation of impedance (Z) of a series LCR circuit vs frequency (f) of the ac supply. Mark the resonant frequency as $f_0$.
Draw: labelled graph showing variation of impedance (Z) of a series LCR circuit vs frequency (f) of the ac supply
Must show: impedance Z on y-axis, frequency f on x-axis, minimum impedance Z = R at resonant frequency f0
Answer
Answer (a)
Formulas used: Inductive reactance: XL = 2*pi*f*L with L = 1 H, f = 50 Hz, so XL = 314.2 ohm Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with R = 400 ohm, XL = 314.2 ohm, so Z = 508.6 ohm Ohm's law: V = I*R with R = 508.6 ohm, V = 220 V, so I = 0.4325 A
(1) Reactance of the coil: $X_L = 2\pi fL = 2\pi\times50\times1 = 314.2\,\Omega$.
(2) Impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{400^2 + 314.2^2} = 508.6\,\Omega$.
(3) Current: $I = \frac{V}{Z} = \frac{220}{508.6} = 0.433\text{ A}$.
Final answer: 0.433 A
Answer (b)
A graph of impedance $Z$ (y-axis) against frequency $f$ (x-axis) for a series LCR circuit. $Z$ falls from a high value at low frequency (capacitive, $X_C$ large) to a minimum value $Z = R$ at the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$, then rises again for $f > f_0$ (inductive). The curve is a U-shape (a dip) with the minimum at $f_0$, and $R$ is marked on the $Z$ axis.
U-shaped Z-f curve with its minimum Z = R at f0 = 1/(2 pi sqrt(LC)); axes impedance Z and frequency f.