Question

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

Answer 1

Given R1 = R2 = 2Ω

Internal resistance = r

In parallel current flows = 1.2 A

In series current flows = 0.4A

Assuming that potential of the source remains constant in both cases.

Equivalent Resistance for Parallel connection

Rp = r1*r2 / r1+r2 = 2x2/2+2 = 1ohm

Thus, sum total of potential drops at individual resistance = total potential difference applied
i(Rp+r) =V

=1.2(1+r)=V--------------1.

Now, for series connection

Equivalent resistance together with the internal resistance in the series it becomes
Rs=r+2+2 =4+r

Thus, sum total of potential drops at individual resistance = total potential difference applied

i(Rs) =V

0.4 x 4+r = V----------------2.

From eqn. 1 and eqn. 2

1.2x(1+r)=0.4x(4+r)
=0.5ohm


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