Question

If $E$ is the emf of a cell of internal resistance $r$. It is connected with an external resistance $R$, then potential difference across $R$ is given as

• A. $V=\frac{E}{(R+r)}$
• B. $V=E$
• C. $V=\frac{E}{(1+r/R)}$
• D. $V=\frac{E}{(1+R/r)}$

Answer 1

The correct option is C $V=\frac{E}{(1+r/R)}$

Circuit with the external resistance:


Current in the circuit, $i=\frac{E}{R+r}$

Now, potential difference across $R$,

$V=iR=\left(\frac{E}{R+r}\right)R$

$V=E\left(\frac{R}{R+r}\right)$

$V=\frac{E}{1+r/R}$

Why this question? Key concept: Battery acting as a source (or battery is discharging) $V_A-V_B=\epsilon -ir$ $V_A-V_B$ is also called terminal voltage.