Question

26. A cell supplies a current of 1.2 A through two
resistors each of 2 12 connected in parallel. When
the resistors are connected in series, it supplies a
current of 0.4 A. Calculate : (i) the internal
resistance, and (ii) e.m.f. of the cell.​

Answer 1

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

Explanation:

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Answer 2

Answer:

Explanation:

1 ST OPTION IS CORRECT I AM THINKING IT IS CORRECT ANSWER