Current Electricity Charging Discharging Of Capacitors Question 442
Question: Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in series. The powers dissipated in them have the ratio
[NCERT 1977]
Options:
A) 1 : 2
B) 2 : 1
C) 1 : 1
D) 1 : 4
Show Answer
Answer:
Correct Answer: A
Solution:
In series, current is same in both the bulbs, hence $ P\propto R,(P=i^{2}R) $
$ \
Therefore \frac{P _{1}}{P _{2}}=\frac{R _{1}}{R _{2}}=\frac{1}{2} $
BETA
