Current Electricity Charging Discharging Of Capacitors Question 129
Question: In the following circuit, 5 W resistor develops 45 J/s due to current flowing through it. The power developed per second across 12 W resistor is
[AMU (Engg.) 1999]
Options:
A) 16 W
B) 192 W
C) 36 W
D) 64 W
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{i _{1}}{i _{2}}=\frac{15}{5}=\frac{3}{1} $ ? (i) Also $ \frac{H}{t}=i^{2}R\Rightarrow 45={{(i _{1})}^{2}}\times 5 $
Therefore $ i _{1}=3,A $ and from equation (i) $ i _{2}=1,A $ So $ i=i _{1}+i _{2}=4,A $ Hence power developed in $ 12,\Omega $ resistance $ P=i^{2}R={{(4)}^{2}}\times 12=192,W $
BETA
