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 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.
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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!
Updated On: Jun 3, 2026
- $0.45 \text{ Vm}^{-1}$
- $0.5 \text{ Vm}^{-1}$
- $0.2 \text{ Vm}^{-1}$
- $0.1 \text{ Vm}^{-1}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Potential gradient ($k$) is defined as the potential drop per unit length along a potentiometer wire: $k = \frac{V_w}{L}$, where $V_w$ is the voltage drop across the wire and $L$ is its total length.
Step 2: Meaning
The voltage drop $V_w$ across the wire can be found using Ohm's Law ($V_w = I \cdot R_w$), where $I$ is the total circuit current and $R_w$ is the resistance of the potentiometer wire.
Step 3: Analysis
First, find the total current in the main circuit loop: $I = \frac{E}{R_w + r} = \frac{5}{18 + 2} = \frac{5}{20} = 0.25 \text{ A}$. Next, calculate the voltage drop across the wire: $V_w = I \times R_w = 0.25 \times 18 = 4.5 \text{ V}$. Finally, divide by the total length of the wire ($L = 10 \text{ m}$) to find the potential gradient: $k = \frac{4.5 \text{ V}}{10 \text{ m}} = 0.45 \text{ Vm}^{-1}$.
Step 4: Conclusion
Thus, the potential gradient along the potentiometer wire is $0.45 \text{ Vm}^{-1}$.
Final Answer: (A)
Potential gradient ($k$) is defined as the potential drop per unit length along a potentiometer wire: $k = \frac{V_w}{L}$, where $V_w$ is the voltage drop across the wire and $L$ is its total length.
Step 2: Meaning
The voltage drop $V_w$ across the wire can be found using Ohm's Law ($V_w = I \cdot R_w$), where $I$ is the total circuit current and $R_w$ is the resistance of the potentiometer wire.
Step 3: Analysis
First, find the total current in the main circuit loop: $I = \frac{E}{R_w + r} = \frac{5}{18 + 2} = \frac{5}{20} = 0.25 \text{ A}$. Next, calculate the voltage drop across the wire: $V_w = I \times R_w = 0.25 \times 18 = 4.5 \text{ V}$. Finally, divide by the total length of the wire ($L = 10 \text{ m}$) to find the potential gradient: $k = \frac{4.5 \text{ V}}{10 \text{ m}} = 0.45 \text{ Vm}^{-1}$.
Step 4: Conclusion
Thus, the potential gradient along the potentiometer wire is $0.45 \text{ Vm}^{-1}$.
Final Answer: (A)
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