Current Electricity Ques 66
- Two equal resistances when connected in series to a battery consume electric power of $60$ $ W$. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be
(2019 Main, 11 Jan I)
(a) $60 $ $W$
(b) $30 $ $W$
(c) $240 $ $W$
(d) $120 $ $W$
Show Answer
Answer:
Correct Answer: 66.(c)
Solution:
Formula:
- Let $P _1$ and $P _2$ be the individual electric powers of the two resistances, respectively.
In series combination, power is
$ P _0=\frac{P _1 P _2}{P _1+P _2}=60$ $ W $
Since, the resistances are equal and the current through each resistor in series combination is also same. Then,
$ P _1=P _2=120 $ $W $
In parallel combination, power is
$ P=P _1+P _2=120+120=240 $ $ W $
Alternate method
Let $R$ be the resistance.
$\therefore$ Net resistance in series $=R+R=2 R$
$ \begin{aligned} P & =\frac{V^{2}}{2 R}=60 W \\ \Rightarrow \quad \frac{V^{2}}{R} & =120 W \end{aligned} $
New resistance in parallel $=\frac{R \times R}{R+R}=R / 2$
$ P^{\prime}=\frac{V^{2}}{R / 2}=2 (\frac{V^{2}}{R})=240 $ $W $
BETA
