Two identical cells each of emf E and internal resistance r are connected in parallel with an external resistance R. To get maximum power developed across R, the value of R is

Question

Two identical cells each of emf E and internal resistance r are connected in parallel with an external resistance R. To get maximum power developed across R, the value of R is (A) R= (r/2)  (B) R = r (C) R= (r/3)  (D) R =2r

Tardigrade Admin · Accepted Answer

Equivalent resistance Req= (r/2)+R= (r+2R/2) ∴ I= (2E/r+2R) For maximum power consumption, I should be maximum so denominator is minimum. For this r+2R=(√ r- √ 2R)2+2 √ r √ 2R ⇒ √ r- √ 2R=0 ⇒ R= (r/2)

Q. Two identical cells each of emf E and internal resistance r are connected in parallel with an external resistance R. To get maximum power developed across R, the value of R is

$R = \frac{r}{2}$

R = r

$R = \frac{r}{3}$

R =2r

Equivalent resistance $R_{e q} = \frac{r}{2} + R = \frac{r + 2 R}{2}$
$∴ I = \frac{2 E}{r + 2 R}$
For maximum power consumption, I should be maximum so denominator is minimum. For this
$r + 2 R = (r - 2 R)^{2} + 2 r 2 R$
$\Rightarrow r - 2 R = 0$
$\Rightarrow R = \frac{r}{2}$

Q. Two identical cells each of emf E and internal resistance r are connected in parallel with an external resistance R. To get maximum power developed across R, the value of R is

R=2r​

R = r

R=3r​

R =2r

Equivalent resistance Req​=2r​+R=2r+2R​ ∴I=r+2R2E​ For maximum power consumption, I should be maximum so denominator is minimum. For this r+2R=(r​−2R​)2+2r​2R​ ⇒r​−2R​=0 ⇒R=2r​

$R = \frac{r}{2}$

$R = \frac{r}{3}$

Equivalent resistance $R_{e q} = \frac{r}{2} + R = \frac{r + 2 R}{2}$
$∴ I = \frac{2 E}{r + 2 R}$
For maximum power consumption, I should be maximum so denominator is minimum. For this
$r + 2 R = (r - 2 R)^{2} + 2 r 2 R$
$\Rightarrow r - 2 R = 0$
$\Rightarrow R = \frac{r}{2}$