Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2(r1r2). If the potential difference across the source of internal resistance r1 is zero then the value of R will be

Question

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2(r1>r2). If the potential difference across the source of internal resistance r1 is zero then the value of R will be (A) r1-r2 (B) ( r 1 r2/ r 1+r2) (C) ( r 1+r2/2) (D) r 2- r 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/33d6cd64fc83be1f5a07638927076ffd-.png" /> I =(2 E / r 1+ r 2+ R ) IR = E - Ir 2 I ( R + r 2)= E I =( E / R + r 2) (2 E / r 1+ r 2+ R )=( E / R + r 2) 2 R +2 r 2= r 1+ r 2+ R R = r 1- r 2

$r_{1} - r_{2}$

$\frac{r _{1} r _{2}}{r _{1} + r _{2}}$

$\frac{r _{1} + r _{2}}{2}$

$r_{2} - r_{1}$

$I = \frac{2 E}{r _{1} + r _{2} + R}$
$I R = E - I r_{2}$
$I (R + r_{2}) = E$
$I = \frac{E}{R + r _{2}}$
$\frac{2 E}{r _{1} + r _{2} + R} = \frac{E}{R + r _{2}}$
$2 R + 2 r_{2} = r_{1} + r_{2} + R$
$R = r_{1} - r_{2}$

r1​−r2​

r1​+r2​r1​r2​​

2r1​+r2​​

r2​−r1​

I=r1​+r2​+R2E​ IR=E−Ir2​ I(R+r2​)=E I=R+r2​E​ r1​+r2​+R2E​=R+r2​E​ 2R+2r2​=r1​+r2​+R R=r1​−r2​

$r_{1} - r_{2}$

$\frac{r _{1} r _{2}}{r _{1} + r _{2}}$

$\frac{r _{1} + r _{2}}{2}$

$r_{2} - r_{1}$

$I = \frac{2 E}{r _{1} + r _{2} + R}$
$I R = E - I r_{2}$
$I (R + r_{2}) = E$
$I = \frac{E}{R + r _{2}}$
$\frac{2 E}{r _{1} + r _{2} + R} = \frac{E}{R + r _{2}}$
$2 R + 2 r_{2} = r_{1} + r_{2} + R$
$R = r_{1} - r_{2}$