A cell of emf E having an internal resistance r is connected to an external resistance R . The potential difference V across the resistance R varies with R as shown by the curve:

Question

A cell of emf E having an internal resistance r is connected to an external resistance R . The potential difference V across the resistance R varies with R as shown by the curve: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8cbaxeqv4nvqbnrw.jpg" /> (A) A (B) B (C) C (D) D

Tardigrade Admin · Accepted Answer

V=E-ir=E(R + r - r/R + r) ⇒ V=(E R/R + r) ⇒ text V = 0 at textR = 0 and V=E at R=∞

Q. A cell of emf $E$ having an internal resistance $r$ is connected to an external resistance $R$ . The potential difference $V$ across the resistance $R$ varies with $R$ as shown by the curve:

A

B

C

D

$V = E - i r = E \frac{R + r - r}{R + r}$
$\Rightarrow V = \frac{ER}{R + r}$
$\Rightarrow V = 0$ at $R = 0$
and $V = E$ at $R = \infty$

Q. A cell of emf E having an internal resistance r is connected to an external resistance R . The potential difference V across the resistance R varies with R as shown by the curve:

A

B

C

D

V=E−ir=ER+rR+r−r​ ⇒V=R+rER​ ⇒ V=0 at R=0 and V=E at R=∞

Q. A cell of emf $E$ having an internal resistance $r$ is connected to an external resistance $R$ . The potential difference $V$ across the resistance $R$ varies with $R$ as shown by the curve:

$V = E - i r = E \frac{R + r - r}{R + r}$
$\Rightarrow V = \frac{ER}{R + r}$
$\Rightarrow V = 0$ at $R = 0$
and $V = E$ at $R = \infty$