JEE PYQ: Alternating Current Question 21
Question 21 - 2020 (03 Sep Shift 1)
A 750 Hz, 20 V (rms) source is connected to a resistance of $100,\Omega$, an inductance of 0.1803 H and a capacitance of 10 $\mu$F all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C is close to:
(a) 418 s (b) 245 s (c) 365 s (d) 348 s
Show Answer
Answer: (d)
Solution
$Z \approx 835\Omega$. $H = \frac{V^2}{Z^2}Rt = ms\Delta T$. $t \approx 348.61$ s.
BETA
