Two batteries each of emf E and internal resistance r are connected turn by turn in series and in parallel, and are used to find current in an external resistance R. If the current in series is equal to that in parallel, the internal resistance of each battery is:

Question

Two batteries each of emf E and internal resistance r are connected turn by turn in series and in parallel, and are used to find current in an external resistance R. If the current in series is equal to that in parallel, the internal resistance of each battery is: (A) R (B) R/2 (C) R/3 (D) R/4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a48ef733f786598544f611b9a13a77dc-.png" /> IS=(2 E/2 r+R) and IP=(E/(r)2+R)=(2 E/r+2 R) If IS=IP (2 E/2 r+R)=(2 E/r+2 R) 2 r+R=r+2 R r=R

Q. Two batteries each of emf $E$ and internal resistance $r$ are connected turn by turn in series and in parallel, and are used to find current in an external resistance $R$ . If the current in series is equal to that in parallel, the internal resistance of each battery is:

R

R/2

R/3

R/4

$I_{S} = \frac{2 E}{2 r + R}$ and $I_{P} = \frac{E}{\frac{r}{2} + R} = \frac{2 E}{r + 2 R}$
If $I_{S} = I_{P}$
$\frac{2 E}{2 r + R} = \frac{2 E}{r + 2 R}$
$2 r + R = r + 2 R$
$r = R$

Q. Two batteries each of emf E and internal resistance r are connected turn by turn in series and in parallel, and are used to find current in an external resistance R. If the current in series is equal to that in parallel, the internal resistance of each battery is:

R

R/2

R/3

R/4

IS​=2r+R2E​ and IP​=2r​+RE​=r+2R2E​ If IS​=IP​ 2r+R2E​=r+2R2E​ 2r+R=r+2R r=R

Q. Two batteries each of emf $E$ and internal resistance $r$ are connected turn by turn in series and in parallel, and are used to find current in an external resistance $R$ . If the current in series is equal to that in parallel, the internal resistance of each battery is:

$I_{S} = \frac{2 E}{2 r + R}$ and $I_{P} = \frac{E}{\frac{r}{2} + R} = \frac{2 E}{r + 2 R}$
If $I_{S} = I_{P}$
$\frac{2 E}{2 r + R} = \frac{2 E}{r + 2 R}$
$2 r + R = r + 2 R$
$r = R$