In a potentiometer, a wire of length $10 \mathrm{~m}$ having resistance $50 \Omega$ is used. A battery of $5…
In a potentiometer, a wire of length $10 \mathrm{~m}$ having resistance $50 \Omega$ is used. A battery of $5 \mathrm{~V}$ and a resistor of $450 \Omega$ are connected in series to the wire. If an unknown battery of emf $E$ balances the potentiometer at $450 \mathrm{~cm}$, then the value of $E$ is
0.225 V
1.25 V
2.25 V
0.0225 V
Solution
Given length of wire, $l=10 \mathrm{~m}$, resistance of wire, $R=50 \Omega$, emf of battery, $E_1=5 \mathrm{~V}$, balancing length, $x=450 \mathrm{~cm}=4.5 \mathrm{~m}$ series resistor, $R_1=450 \Omega$ Current, $i=\frac{E_1}{R+R_1}=\frac{5}{50+450}=\frac{5}{500}=0.01 \mathrm{~A}$ So, $\quad V=i R=.01 \times 50=0.5 \mathrm{~V}$ emf of primary cell,
$
E=\frac{V_x}{l}=\frac{0.5 \times 4.50}{10}=0.225 \mathrm{~V}
$