Question

A potentiometer wire is 10m long and has a resistance of $18\,\Omega$. It is connected to a battery of emf 5V and internal resistance $2\,\Omega$. Calculate the potential gradient along the wire.

• A. $0.45 \text{ Vm}^{-1}$
• B. $0.5 \text{ Vm}^{-1}$
• C. $0.2 \text{ Vm}^{-1}$
• D. $0.1 \text{ Vm}^{-1}$

Hint:

Always remember to include the internal resistance ($r$) when calculating the total current: $I = \frac{E}{R_w + r}$. Don't just divide the battery voltage by the wire resistance!

Answer 1

The Correct Option is A

Step 1: What is potential gradient?
It is the voltage drop per metre along the wire: $k = \dfrac{V_{w}}{L}$.

Step 2: Find the circuit current.
The wire ($18\,\Omega$) and internal resistance ($2\,\Omega$) are in series with the $5$ V cell. \[ I = \frac{E}{R_{w}+r} = \frac{5}{18+2} = 0.25 \text{ A} \]
Step 3: Voltage across the wire.
\[ V_{w} = I\,R_{w} = 0.25\times 18 = 4.5 \text{ V} \]
Step 4: Note the wire length.
The wire is $L = 10$ m long.

Step 5: Divide to get the gradient.
\[ k = \frac{4.5}{10} = 0.45 \text{ V m}^{-1} \]
Step 6: State the answer.
So the potential gradient is $0.45$ V m$^{-1}$, which is option 1.
\[ \boxed{k = 0.45 \text{ V m}^{-1}} \]