Current Electricity Charging Discharging Of Capacitors Question 92
Question: In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is 60 calorie per second. The heat generated across 3 ohm resistance per second will be
[MP PET 1996]
Options:
A) 30 calories
B) 60 calories
C) 100 calories
D) 120 calories
Show Answer
Answer:
Correct Answer: D
Solution:
Resistance of upper branch $ R _{1}=2+3=5,\Omega $ Resistance of lower branch $ R _{2}=4+6=10,\Omega $ Hence $ \frac{i _{1}}{i _{2}}=\frac{R _{2}}{R _{1}}=\frac{10}{5}=2 $
$ \frac{\text{Heat generated across 3 }\Omega \text{ (}{{\text{H}} _{\text{1}}})}{\text{Heat generated across 6 }\Omega \text{ (}{{\text{H}} _{\text{2}}})} $
$ =\frac{i _{1}^{2}\times 3}{i _{2}^{2}\times 6}=\frac{4}{2}=2 $ \ Heat generated across 3 Ω = 120 cal/sec
BETA
