Question

Two identical cells of emf 1.5V each joined in parallel supply energy to an external circuit consisting of two resistances of 7Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4V. Calculate the internal resistance of each cell.

Answer 1

Explanation:

refer the pic for answer

Answer 2

The internal resistance of both cell are same as they are identical and it  is 0.50 Ω

1. The total external resistance of the circuit is given by parallel combination
2. $Rp=\frac{R1XR2}{R1+R2} =\frac{7X7}{7+7}=49/14$ =3.5 Ω
3. The internal resistance of circuit is
4. $r=R(\frac{E-V}{V})$ here E is the emf of the cell and V is the terminal voltage
5. r=3.5[(1.5-1.4)/1.4]=0.25 Ω
6. let r1 and r2 is the two internal resitance which is also connected in parallel r1=r2 so
7. $r=\frac{r1Xr2}{r1+r2} =\frac{r1Xr1}{r1+r1}=\frac{r1}{2}$ r1 =2r= 2 x 0.25 =0.50 Ω