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ICSE class X question...Please give the solution....if u are right I will surely mark you BRAINLIEST....if u spam I will report.​

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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


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