Question

a cell supplies a current of 1.2 A through 2ohm resistors connected in parallel.when the resistors are connected in series it supplies a current of 0.4A.calculate the internal resistance and e.m.f of the cell

Answer 1

HEY DEAR .

-----------------------------------------

-----------------------------------------


Given=R1=R2=2ohms
Internal resistance=r
In parallel current flows=1.2 A
In series current flows =0.4A
Assuming potential of source remains  constant in both case:
Equivalent resistant in parallel :
Rp=R1R2/R1+R2
=2x2/2+2
=1ohm
Thus , sum of potential drops at individual resistance=total potential difference applied
i(Rp+r)=v
1.2(1+r)=V-----------------1
now for series connection:
Rs=2+2+r=r+4
Thus sum of potential drops at individual resistance=total potential difference applied
iRs=V
0.4x(4+r)=V------------------2
from eq.  1 and 2
1.2(1+r)=0.4x(4+r)
r=0.5ohms
on substituting r=0.5 ohms in equ 1 we get E=1.8volts

HOPE , IT HELPS . ✌