Question

8.
Five cells each of internal resistance 0.2ohm
and e.m.f. 2 V are connected in series with a
resistance of 4 ohm. The current through the
external resistance is
(a) 4 A '(b) 2 A (c)1 A '(d) 0.5 A​

Answer 1

Answer:

b. 2A

Explanation:

I=E/r+R

E of 5 cell is 2×5=10v

r of 5 cell is 0.2×5=1ohm

R is 4 ohm

I =10/1+4 = 10/5

=2A