If the potential difference across the internal resistance r1 is equal to the emf E of the battery, then

Question

If the potential difference across the internal resistance r1 is equal to the emf E of the battery, then (A) R = r1 +r2 (B) R = (r1/r2) (C) R = r1 -r2 (D) R = (r2/r1)

Tardigrade Admin · Accepted Answer

The total emf =E+E=2 E Total resistance =R+r1+r2 ∴ Current flowing through the circuit i=(2 E/R+r1+r2) According to question E=i r1 ⇒ i=(E/r1) ∴ (E/r1)=(2 E/R+r1+r2) R=r1-r2

Q. If the potential difference across the internal resistance $r_{1}$ is equal to the emf $E$ of the battery, then

$R = r_{1} + r_{2}$

$R = \frac{r _{1}}{r _{2}}$

$R = r_{1} - r_{2}$

$R = \frac{r _{2}}{r _{1}}$

The total $e m f = E + E = 2 E$
Total resistance $= R + r_{1} + r_{2}$
$∴$ Current flowing through the circuit
$i = \frac{2 E}{R + r _{1} + r _{2}}$
According to question $E = i r_{1}$
$\Rightarrow i = \frac{E}{r _{1}}$
$∴ \frac{E}{r _{1}} = \frac{2 E}{R + r _{1} + r _{2}}$
$R = r_{1} - r_{2}$

Q. If the potential difference across the internal resistance r1​ is equal to the emf E of the battery, then

R=r1​+r2​

R=r2​r1​​

R=r1​−r2​

R=r1​r2​​