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Solve the following.
An alternating emf E = 5.0 sin(314t) V is applied to a circuit containing a resistor connected in…
An alternating emf $E = 5.0 \sin(314t)\text{ V}$ is applied to a circuit containing a resistor connected in series with an unknown component X.
The current in the circuit is found to be $I = 3.0 \sin\left(314t - \frac{\pi}{3}\right)\text{ A}$.
(a)[1]
Identify the component X.
(b)[1]
Calculate rms value ($I_{\text{rms}}$) of the current flowing through the circuit.
(c)[1]
Find the frequency of the source.
(d)[1]
Calculate power factor.
(e)[1]
Find the impedance ($Z$) of the circuit.
Answer
Answer (a)
X is an inductor, because the current lags the emf by $\frac{\pi}{3}$.
Answer (b)
Formula used: rms and peak values: rms = peak/sqrt(2) with peak = 3 A, so rms = 2.12 A
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{3.0}{\sqrt{2}} = 2.12\text{ A}$.
Final answer: 2.12 A
Answer (c)
Comparing with $E = E_0\sin\omega t$: $\omega = 314\text{ rad s}^{-1}$, so $f = \frac{\omega}{2\pi} = \frac{314}{2\pi} = 50\text{ Hz}$.
Final answer: 50 Hz
Answer (d)
Phase difference $\phi = \frac{\pi}{3} = 60^\circ$, so power factor $= \cos\phi = \cos60^\circ = 0.5$.
Final answer: 0.5
Answer (e)
Formula used: Ohm's law: V = I*R with I = 3 A, V = 5 V, so R = 1.67 ohm
$Z = \frac{E_0}{I_0} = \frac{5.0}{3.0} \approx 1.67\ \Omega$.
Final answer: 1.67 ohm