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Solve the following.
An 8 H inductor, a 2 μF capacitor and a 100 Ω resistor are connected in series to an A.C. supply of…
An $8\text{ H}$ inductor, a $2\ \mu\text{F}$ capacitor and a $100\ \Omega$ resistor are connected in series to an A.C. supply of $220\text{ V}$ and $50\text{ Hz}$. Calculate:
(i)[1]
Impedance of the circuit.
(ii)[1]
Current flowing through the circuit.
(iii)[1.5]
Phase difference between the current and the supply voltage.
(iv)[1.5]
Average power consumed by the circuit.
Answer
Answer (i)
Formula used: Series LCR circuit from L, C and f: Z = sqrt(R**2 + (XL - XC)**2) with C = 2 uF, L = 8 H, R = 100 ohm, f = 50 Hz, so Z = 927 ohm
Given: $L = 8\text{ H}$, $C = 2\ \mu\text{F}$, $R = 100\ \Omega$, $V = 220\text{ V}$, $f = 50\text{ Hz}$.
$X_L = 2\pi f L = 2\pi\times50\times8 = 2513\ \Omega$; $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi\times50\times2\times10^{-6}} = 1592\ \Omega$.
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 921.7^2} \approx 927\ \Omega$.
Final answer: 927 ohm
Answer (ii)
Formula used: Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with V = 220 V, Z = 927 ohm, so I = 0.237 A
$I = \frac{V}{Z} = \frac{220}{927} = 0.237\text{ A}$.
Final answer: 0.237 A
Answer (iii)
Formula used: Phase angle of a series LCR circuit (current lags by phi): phi = atan((XL - XC)/R) with R = 100 ohm, XC = 1592 ohm, XL = 2513 ohm, so phi = 83.8 deg
$\tan\phi = \frac{X_L - X_C}{R} = \frac{921.7}{100} = 9.217$, so $\phi = 83.8^\circ$ (the current lags behind the voltage, as the circuit is inductive).
Final answer: 83.8 degree
Answer (iv)
Formula used: Electric power: P = V*I with I = 0.2373 A, R = 100 ohm, so P = 5.63 W
Average power $P = I^2R = (0.237)^2\times100 = 5.63\text{ W}$ (equivalently $VI\cos\phi$).
Final answer: 5.63 W