Discussion Forum



1)

Two sources of equal emf are connected to a  resistance R. The internal resistance of these sources are r1 and r2  (r1  > r2). if the potential difference across the source having internal resistance r2 is zero, then 


A) $R= \frac{r_{1}r_{2}}{r_{2}-r_{1}}$

B) $R= r_{2}\left(\frac{r_{1}+r_{2}}{r_{2}-r_{1}}\right)$

C) $R= \left(\frac{r_{1}r_{2}}{r_{2}+r_{1}}\right)$

D) R= $r_{2}-r_{1}$

Answer:

Option D

Explanation:

 Let  emf of each source be E . When they are connected in series , the current in the circuit 

$I= \frac{E_{tot}}{R_{tot}}=\frac{E+E}{r_{1}+r_{2}+R}=\frac{2E}{r_{1}+r_{2}+R}$

 potential  drop across the cell of internal

resistance   $r_{2},\left(\frac{2E}{r_{1}+r_{2}+R}\right)r_{2}$

 Hence,  $E=\frac{2E}{(r_{1}+r_{2}+R)}.r_{2}=0$

$r_{1}+r_{2}+R=2r_{2}$

 $\Rightarrow$  $R=r_{2}-r_{1}$