Current Electricity Charging Discharging Of Capacitors Question 485
Question: A coil develops heat of 800 cal/sec. When 20 volts is applied across its ends. The resistance of the coil is (1 cal = 4.2 joule)
[MP PET 1994]
Options:
A) $ 1.2,\Omega $
B) $ 1.4,\Omega $
C) $ 0.12,\Omega $
D) $ 0.14,\Omega $
Show Answer
Answer:
Correct Answer: C
Solution:
$ H=\frac{V^{2}t}{4.2,R} $ or $ \frac{H}{t}=\frac{V^{2}}{4.2,R} $
Therefore $ 800=\frac{20\times 20}{4.2\times R} $
Therefore $ R=\frac{5}{42}=0.119\approx 0.12,\Omega $
BETA
