A cell supplies current of 1.2 A through two resistors each of 2 ohm c

1.	A cell supplies current of 1.2 A through two resistors each of 2 ohm connected in parallel. When the resistors are connected in series it supplies a current of 0.4 A , Calculate : internal resistence and emf of the cell....
Answer» Answer: 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

A cell supplies current of 1.2 A through two resistors each of 2 ohm connected in parallel. When the resistors are connected in series it supplies a current of 0.4 A , Calculate : internal resistence and emf of the cell....

Answer»

Answer:

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