The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

Question

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be (A)  (E2/r) (B)  (E2/2r) (C)  (E2/3r) (D)  (E2/4r)

Tardigrade Admin · Accepted Answer

Current i = (E/r + R) Power p = i2 R ⇒ p = (E2R/(r + R)2) Power will be maximum when r = R pmax = (E2/4r)

Q. The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

$\frac{E ^{2}}{r}$

$\frac{E ^{2}}{2 r}$

$\frac{E ^{2}}{3 r}$

$\frac{E ^{2}}{4 r}$

Current $i = \frac{E}{r + R}$
Power $p = i^{2} R$
$\Rightarrow p = \frac{E ^{2} R}{( r + R ) ^{2}}$
Power will be maximum when r = R
$p_{ma x} = \frac{E ^{2}}{4 r}$

Q. The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

rE2​

2rE2​

3rE2​

4rE2​

Current i=r+RE​ Power p=i2R ⇒p=(r+R)2E2R​ Power will be maximum when r = R pmax​=4rE2​

$\frac{E ^{2}}{r}$

$\frac{E ^{2}}{2 r}$

$\frac{E ^{2}}{3 r}$

$\frac{E ^{2}}{4 r}$

Current $i = \frac{E}{r + R}$
Power $p = i^{2} R$
$\Rightarrow p = \frac{E ^{2} R}{( r + R ) ^{2}}$
Power will be maximum when r = R
$p_{ma x} = \frac{E ^{2}}{4 r}$