Question

When 10 cells in series are connected to the ends of a resistance of 592, the current is found to be 0.25 A, but when the same cells after being connected in parallel are joined to the ends of a 0.052, the current is 25 A. Calculate the internal resistance and emf of each cell. (Ans. 0.112, 1.5 V)​

Answer 1

Answer:

Let emf of each cell =E and internal resistance =r.

For series: E+E=1(r+r+5) (using V=IR)

or 2E=2r+5....(1)

For parallel: E=0.8R

eq

=8(r∣∣r+5)=0.8(

r+r

rr

+5)=0.8(r/2+5)

E=0.4r+4....(2)

From (1) and (2), 0.8r+8=2r+5⇒r=

1.2

3

=5/2=2.5Ω