The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5

Q: Why did we completely ignore the $20 \ \Omega$ internal resistance of the cell?

At the null point, the galvanometer shows zero deflection, meaning zero current flows through the secondary circuit. The voltage across the terminals of a cell is given by $V = E - ir$. Since $i = 0$, the terminal voltage $V$ is exactly equal to the EMF $E$. The internal resistance causes no voltage drop and is essentially irrelevant at the balance point.

Q: Shouldn't we convert the lengths from centimeters to meters?

You can, but it's not strictly necessary as long as you are consistent. By keeping the length in cm, we calculated the resistance per unit length $\lambda$ in $\Omega/\text{cm}$. When we multiplied it by the total length in cm, the units canceled out perfectly to give Ohms. Converting to meters would just add extra powers of 10 that eventually cancel out anyway.

Q: What if the primary current wasn't given directly?

If the primary current $I$ wasn't given, the problem would have to provide the voltage of the primary driving battery and any series resistance connected to the potentiometer wire. You would then calculate $I$ using Ohm's law for the primary loop: $I = \frac{V_{\text{primary}}}{R_{\text{wire}} + R_{\text{series}}}$.

Q: Why is a potentiometer considered better than a standard voltmeter for measuring EMF?

A standard voltmeter draws a small amount of current from the circuit to deflect its pointer, which slightly lowers the voltage it's trying to measure ($V = E - ir$). A potentiometer, at the null point, draws absolutely zero current. It acts like an ideal voltmeter with infinite resistance, giving a perfectly accurate reading of the true EMF.

Q: What would happen if the cell's EMF was $10 \text{ V}$ instead of $5 \text{ V}$ in this exact setup?

Let's check the maximum potential drop across the entire wire. $V_{\text{max}} = I \cdot R = (60 \times 10^{-3} \text{ A}) \times 100 \ \Omega = 6 \text{ V}$. Since the maximum voltage the wire can balance is $6 \text{ V}$, a $10 \text{ V}$ cell could never be balanced. You would not find a null point anywhere on the wire; the galvanometer would always deflect in one direction.

Question

The length of a potentiometer wire is $1200 	ext{ cm}$ and it carries a current of $60 	ext{ mA}$. For a cell of emf $5 	ext{ V}$ and internal resistance of $20 \ \Omega$, the null point on it is found to be at $1000 	ext{ cm}$. The resistance of whole wire is

Options:

(A) $80 \ \Omega$

(B) $100 \ \Omega$

(C) $60 \ \Omega$

(D) $120 \ \Omega$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a long potentiometer wire, $1200 	ext{ cm}$ in length, stretched out on a board. A steady primary current of $60 	ext{ mA}$ is flowing through it, creating a uniform potential gradient along its length.

Now, we connect a secondary circuit containing a $5 	ext{ V}$ cell and a galvanometer. We slide the jockey along the wire and find the null point exactly at $1000 	ext{ cm}$. The question asks us to find the total resistance of this $1200 	ext{ cm}$ wire.

### The Master Equation
What does the null point actually mean? It means the galvanometer shows zero deflection, indicating that absolutely no current is being drawn from the $5 	ext{ V}$ cell. Therefore, the potential drop across the $1000 	ext{ cm}$ segment of the wire perfectly balances the cell's EMF.

Notice how the internal resistance of $20 \ \Omega$ doesn't matter here! Since no current flows through the cell circuit at the null point, there is no voltage drop across its internal resistance ($V = E - ir = E - 0 = E$). This was a classic distractor designed to test your conceptual clarity.

Let's translate this into an equation. If $\lambda$ is the resistance per unit length of our wire, then the resistance of the balancing portion is simply $\lambda \cdot l$. By Ohm's law, the voltage drop across it is the primary current $I$ times this resistance. So, our master equation becomes:

$$E = I \cdot \lambda \cdot l$$

### Finding the Potential Gradient
Now, let's carefully plug in our known values. The EMF $E$ is $5 	ext{ V}$. The primary current $I$ is $60 	ext{ mA}$, which we must write as $60 	imes 10^{-3} 	ext{ A}$ to keep our units standard. The balancing length $l$ is $1000 	ext{ cm}$. We will keep the length in centimeters for now to find $\lambda$ in $\Omega/	ext{cm}$.

$$5 = (60 	imes 10^{-3}) \cdot \lambda \cdot 1000$$

Let's simplify this. The $10^{-3}$ and $1000$ cancel each other out perfectly. We are left with:

$$5 = 60 \cdot \lambda$$

Dividing both sides by $60$, we get:

$$\lambda = \frac{5}{60} = \frac{1}{12} \ \Omega/	ext{cm}$$

This is the resistance of just one centimeter of our wire.

### Final Calculation
We are almost there. The question asks for the total resistance of the entire wire. Since we know the resistance of one centimeter, finding the total resistance is straightforward. We just multiply this $\lambda$ by the total length of the wire, $L$.

$$R = \lambda \cdot L$$

Substitute $\lambda = \frac{1}{12} \ \Omega/	ext{cm}$ and the total length $L = 1200 	ext{ cm}$:

$$R = \frac{1}{12} 	imes 1200$$

$$R = 100 \ \Omega$$

And there we have it! The resistance of the whole wire is **$100 \ \Omega$**. This perfectly matches option (b). A beautiful and straightforward application of the potentiometer principle.

Animated Solution for Physics - Current Electricity: The length of a potentiometer wire is $1200 cm$ and it carries a current of $60 mA$ . For a cell of emf $5 V$ and internal resistance of $20 Ω$ , the null point on it is found to be at $1000 cm$ . The resistance of whole wire is

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Visualized Solution

$Potentiometer Setup$

$Principle of Potentiometer$

$Potential Drop Equation$

$Substituting Values$

$Calculating Resistance per Unit Length$

$Total Resistance of the Wire$

$Final Calculation$

$Conclusion$

$Conceptual Takeaway$

The Sigma Insight: Electrical Instruments

Analyzing the Setup Imagine a long potentiometer wire, $1200 cm$ in length, stretched out on a board

The Master Equation What does the null point actually mean? It means the galvanometer shows zero deflection, indicating that absolutely no current is being drawn from the $5 V$ cell

Finding the Potential Gradient Now, let's carefully plug in our known values

Final Calculation We are almost there

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potentiometer Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Potentiometer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Drop Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Resistance per Unit Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Resistance of the Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Takeaway