Two cells of emf 2 E and E with internal resistance r 1 and r 2 respectively are connected in series to an extemal resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

Question

Two cells of emf 2 E and E with internal resistance r 1 and r 2 respectively are connected in series to an extemal resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3355824957aa54a3ecefe7d1c40fd044-.png" /> (A) r 1+ r 2 (B) (r1/2)-r2 (C) ( r 1/2)+ r 2 (D) r 1- r 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/97d43ba81da38e59d330c97e7612181c-.png" /> i =(3 E / R + r 1+ r 2) TPD =2 E - ir 1=0 2 E = ir 1 2 E =(3 E × r 1/ R + r 1+ r 2) 2 R +2 r 1+2 r 2=3 r 1 R =( r 1/2)- r 2

Q. Two cells of emf $2 E$ and $E$ with internal resistance $r_{1}$ and $r_{2}$ respectively are connected in series to an extemal resistor $R$ (see figure). The value of $R$ , at which the potential difference across the terminals of the first cell becomes zero is

$r_{1} + r_{2}$

$\frac{r _{1}}{2} - r_{2}$

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

$r_{1} - r_{2}$

$i = \frac{3 E}{R + r _{1} + r _{2}}$
$TP D = 2 E - i r_{1} = 0$
$2 E = i r_{1}$
$2 E = \frac{3 E \times r _{1}}{R + r _{1} + r _{2}}$
$2 R + 2 r_{1} + 2 r_{2} = 3 r_{1}$
$R = \frac{r _{1}}{2} - r_{2}$

Q. Two cells of emf 2E and E with internal resistance r1​ and r2​ respectively are connected in series to an extemal resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

r1​+r2​

2r1​​−r2​

2r1​​+r2​

r1​−r2​