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Solve the following.
When an alternating emf e = 310 sin(100π t) V is applied to a series LCR circuit, current flowing…
(i)[3]
When an alternating emf $e = 310 \sin(100\pi t)\text{ V}$ is applied to a series LCR circuit, current flowing through it is $i = 5 \sin(100\pi t + \pi/3)\text{ A}$.
(1) What is the phase difference between the current and the emf?
(2) Calculate the average power consumed by the circuit.
(ii)[2]
Obtain an expression for the resonant frequency ($f_0$) of a series LCR circuit.
Given: At resonance in a series LCR circuit, $X_L = X_C$
To show: $f_0 = \frac{1}{2\pi\sqrt{LC}}$
Answer
Answer (i)
Formula used: Average power in an AC circuit (rms values): P = V*I*pf with I = 3.5355 A, V = 219.2 V, pf = 0.5, so P = 387.5 W
Given: $e=310\sin(100\pi t)$, $i=5\sin(100\pi t+\pi/3)$, so $E_0=310\ \text{V}$, $I_0=5\ \text{A}$.
(1) Phase difference between current and emf $\phi=\frac{\pi}{3}$ ($60^\circ$), the current leading the emf.
(2) $P_{av}=E_{rms}I_{rms}\cos\phi=\frac{310}{\sqrt2}\times\frac{5}{\sqrt2}\times\cos60^\circ=775\times0.5\ \text{W}$
Final answer: $\phi = \pi/3,\ P = 387.5\ \text{W}$
Answer (ii)
Derivation of the resonant frequency of a series LCR circuit.
- At resonance the current is maximum and the impedance is minimum, which happens when $X_L=X_C$ (the reactances cancel and $Z=R$).
- Substitute $X_L=\omega_0L$ and $X_C=\frac{1}{\omega_0C}$: $\omega_0L=\frac{1}{\omega_0C}$.
- Hence $\omega_0^2=\frac{1}{LC}$, so $\omega_0=\frac{1}{\sqrt{LC}}$.
- With $\omega_0=2\pi f_0$: $f_0 = \frac{1}{2\pi\sqrt{LC}}$.