Answer the following questions: The graph below shows variation of current (I) flowing through an…
Answer the following questions:
(a)
The graph below shows variation of current (I) flowing through an electrical device with time (t), when an ac source $e = 240\sin(\omega t)\text{ V}$ is connected to it.
(1) Identify the device. (resistor, capacitor or an inductor)
(2) Calculate the frequency of the current flowing through the device.
(3) How much is the opposition offered by this device?
(b)
A train is moving on horizontal tracks that are $1\cdot2\text{ m}$ apart. Its displacement (x) depends on time (t) as $x(\text{m}) = 3t^2 + 6$. Vertical component of earth’s magnetic field in its region of motion is $2 \times 10^{-5}\text{ T}$. Calculate the emf induced between two ends of its axle rod $10\text{ s}$ after the train starts moving.
Answer
Answer (a)
AIWritten by AI (gemini) - it can contain mistakes.
(1) Capacitor (the current leads the applied voltage by $90^\circ$ or $\pi/2$).
(2) Time period $T = 80\text{ ms} = 80 \times 10^{-3}\text{ s}$. Frequency $f = \frac{1}{T} = \frac{1}{80 \times 10^{-3}} = 12\cdot5\text{ Hz}$.
(3) Opposition is capacitive reactance $X_C = \frac{e_0}{i_0} = \frac{240}{6} = 40\,\Omega$.
Answer (b)
AIWritten by AI (gemini) - it can contain mistakes.
$x = 3t^2 + 6 \implies v = \frac{dx}{dt} = 6t$. At $t = 10\text{ s}$, $v = 6 \times 10 = 60\text{ m s}^{-1}$.
Induced emf $\mathcal{E} = B_v L v = 2 \times 10^{-5} \times 1\cdot2 \times 60 = 1\cdot44 \times 10^{-3}\text{ V} = 1\cdot44\text{ mV}$.
Final answer: 1.44e-3 V
From ISC 2027 Specimen Physics Paper 1, question 18(i).