Current Electricity Ques 32
32. A set of ’ $n$ ’ equal resistors, of value ’ $R$ ’ each, are connected in series to a battery of emf ‘E’ and internal resistance ’ $R$ ‘. The current drawn is I. Now, the ’n’ resistors are connected in parallel to the same battery. Then the current drawn from battery becomes $10 I$. The value of ’n’ is
[2018]
(a) 10
(b) 11
(c) 9
(d) 20
Show Answer
Answer:
Correct Answer: 32.(a)
Solution:
- (a) In series grouping equivalent resistance $R _{\text{series }}=nR$
In parallel grouping equivalent resistance
$R _{\text{parallel }}=\frac{R}{n}$
$I=\frac{E}{n R+R}\quad$ ….(i)
$10 I=\frac{E}{\frac{R}{n}+R}\quad$ ….(ii)
Dividing eq. (ii) by (i),
$ 10=\frac{(n+1) R}{(\frac{1}{n}+1) R} $
Solving we get, $n=10$
BETA
