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Solve the following.
An 80 Ω resistor, a 1·0 H inductor and a 40 μF capacitor are connected in series to an a.c. supply…
An $80\,\Omega$ resistor, a $1\cdot0\text{ H}$ inductor and a $40\,\mu\text{F}$ capacitor are connected in series to an a.c. supply of $220\text{ V}$ such that the circuit draws the maximum current. At this stage, calculate:
(a)[1]
Resonant frequency of the circuit.
(b)[1]
Capacitive reactance of the circuit.
(c)[1]
Impedance of the circuit
(d)[1]
Current flowing through the circuit.
(e)[1]
Phase difference between capacitive and inductive voltages.
Answer
Answer (a)
Formula used: Resonant frequency: f0 = 1/(2*pi*sqrt(L*C)) with C = 40 uF, L = 1.0 H, so f0 = 25.2 Hz
At maximum current the circuit is at resonance: $f_0=\frac{1}{2\pi\sqrt{LC}}=\frac{1}{2\pi\sqrt{1.0\times40\times10^{-6}}}=25.2\text{ Hz}$.
Final answer: 25.2 Hz
Answer (b)
Formula used: Capacitive reactance: XC = 1/(2*pi*f*C) with C = 40 uF, f = 25.1646 Hz, so XC = 158 ohm
$X_C=\frac{1}{2\pi f_0C}=\frac{1}{2\pi\times25.16\times40\times10^{-6}}\approx158\,\Omega$ (equal to $X_L$ at resonance).
Final answer: 158 ohm
Answer (c)
At resonance $X_L=X_C$, so $Z=\sqrt{R^2+(X_L-X_C)^2}=R=80\,\Omega$.
Final answer: 80 ohm
Answer (d)
Formula used: Ohm's law: V = I*R with R = 80 ohm, V = 220 V, so I = 2.75 A
$I=\frac{V}{Z}=\frac{220}{80}=2.75\text{ A}$.
Final answer: 2.75 A
Answer (e)
The voltage across the inductor leads the current by $90^\circ$ and the voltage across the capacitor lags it by $90^\circ$, so the phase difference between them is $180^\circ$ ($\pi$ rad).
Final answer: 180 degree