Question

A voltmeter of resistance $280\,\Omega $ reads the voltage across the terminals of an old dry cell to be $1.40\, V$, while a potentiometer reads its voltage equal to $1.55 \,V$. To draw maximum power from the battery, the load resistance must have the value

• A. $ 60\,\Omega$
• B. $ 45\,\Omega $
• C. $ 35\,\Omega $
• D. $ 30\,\Omega $

Answer 1

Answer: (D)

Let $\varepsilon$ be emf and $r$ be internal resistance of the cell When no current drawn from the cell, the potential difference between its terminals is the emf of the cell It is this quantity which the potentiometer measures
Therefore, emf of the cell, $\varepsilon=1.55\, V$
When the $280\, \Omega$ voltmeter is connected across the cell as shown in adjacent figure,
current flows from the cell The potential difference between its terminals is just the voltmeter reading, $1.40\, V$
$\therefore I \times 280=1.40$
$I=\frac{1.4}{280}=5 \times 10^{-3} A$
Also $I=\frac{\varepsilon}{280+r}$
$=\frac{1.55}{280+r}$
or $5 \times 10^{-3}$
$=\frac{1.55}{280+r}$
$1.4+5 \times 10^{-3} r=1.55$
or $5 \times 10^{-3} r=0.15$
$r=\frac{0.15}{5 \times 10^{-3}}$
$=30\, \Omega$
According to maximum power theorem,
Load resistance, $R =r=30\, \Omega$