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MARCH 2024 — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsMARCH 20243 Marks
Question
An alternating voltage is given by $e = 8sin628.4t.$ Find
$1.$ peak value of e.m.f.
$2.$ frequency or e.m.f.
$3.$ instantaneous value of e.m.f. at time $t = 10.$ ms
✓

Answer

Given an alternating voltage $e = 8sin⁡(628.4t),$ we can analyze it as follows:
$1.$ Peak value of e.m.f. $(E_0​):$ The coefficient in front of the sine function gives us the peak value of the e.m.f., which is $8$ volts in this case.
$2.$ Frequency of e.m.f.: The angular frequency $ω$ is given as $628.4 rad/s.$ The frequency $f$ can be found using the
relationship $\omega=2 \pi f$, so $f=\frac{\omega}{2 \pi}$.
Let's calculate $f:$
$f=\frac{628.4}{2 \pi}$
$3.$ Instantaneous value of e.m.f. at time $t = 10 ms:$ To find the instantaneous value of e.m.f. at $t = 10 ms (0.01 s),$ we substitute t into the given equation for e.
$e(t) = 8sin(628.4 \times 0.01)$
Let's calculate the frequency and the instantaneous value of e.m.f.:
$1.$The peak value of e.m.f. is approximately $8$ volts.
$2.$The frequency of the e.m.f. is approximately $100 Hz.$
$3.$The instantaneous value of e.m.f. at time $t = 10 \ ms$ is approximately $0.0065$ volts.

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