The current in the primary circuit of a potentiometer is 0.2 A. The specific resistance and cross-se

Q: Why didn't they give us the length of the potentiometer wire?

Because you don't need it! When you express the voltage drop $V$ as $I \times \frac{\rho L}{A}$ and divide it by $L$ to find the potential gradient, the length $L$ cancels out completely. The potential gradient depends only on the current, resistivity, and cross-sectional area.

Q: What does 'specific resistance' mean?

Specific resistance is just another term for **resistivity** ($\rho$). It is an intrinsic property of the material the wire is made of, representing how strongly it resists the flow of electric current.

Q: Can the potential gradient change along the wire?

In an ideal potentiometer, the wire is uniform (constant area $A$ and resistivity $\rho$) and the current $I$ is steady. Therefore, the potential gradient $K = \frac{I\rho}{A}$ remains perfectly constant everywhere along the wire.

Q: What if the current in the primary circuit fluctuates?

If the primary current $I$ fluctuates, the potential gradient $K$ will also fluctuate proportionally. This is why a highly stable voltage source (like a lead-acid accumulator) is used in the primary circuit of a potentiometer.

Q: Why is the potential gradient important in a potentiometer?

The potential gradient is the 'calibration factor' of the potentiometer. Once you know $K$, you can find any unknown potential difference simply by finding the balancing length $l$ and multiplying it by $K$ ($V_{unknown} = K \times l$).

Question

The current in the primary circuit of a potentiometer is $0.2	ext{ A}$. The specific resistance and cross-section of the potentiometer wire are $4 	imes 10^{-7}\ \Omega	ext{-m}$ and $8 	imes 10^{-7}\ 	ext{m}^2$, respectively. The potential gradient will be equal to

Options:

(A) $0.2	ext{ V/m}$

(B) $1	ext{ V/m}$

(C) $0.3	ext{ V/m}$

(D) $0.1	ext{ V/m}$

Sigma Percentile · Accepted Answer

### The Essence of a Potentiometer

Imagine a potentiometer wire connected to a primary circuit. A steady current of $0.2	ext{ A}$ flows through this wire, which has a specific resistance (resistivity) $ho = 4 	imes 10^{-7}\ \Omega	ext{-m}$ and a cross-sectional area $A = 8 	imes 10^{-7}\ 	ext{m}^2$. Our goal is to find the **potential gradient** of this wire.

### What is Potential Gradient?

The potential gradient, often denoted by $K$, is simply the voltage drop across the wire per unit length. Mathematically, it is expressed as:

$$K = \frac{V}{L}$$

where $V$ is the total potential difference across the wire and $L$ is its total length.

### Connecting Ohm's Law

According to Ohm's law, the voltage $V$ across the wire is the product of the current $I$ and its resistance $R$:

$$V = IR$$

We also know that the resistance $R$ of a uniform wire depends on its material and dimensions:

$$R = \frac{ho L}{A}$$

Substituting this expression for $R$ into our voltage equation gives:

$$V = I \left(\frac{ho L}{A}ight)$$

### The Master Formula

Now, let's substitute this expression for $V$ back into our potential gradient formula:

$$K = \frac{I \left(\frac{ho L}{A}ight)}{L}$$

Notice how the length $L$ beautifully cancels out! We are left with a neat, length-independent formula for the potential gradient:

$$K = \frac{I ho}{A}$$

This tells us that as long as the current, material, and thickness of the wire are constant, the potential gradient remains uniform throughout the wire.

### Final Calculation

We have all the necessary values to compute $K$:
- $I = 0.2	ext{ A}$
- $ho = 4 	imes 10^{-7}\ \Omega	ext{-m}$
- $A = 8 	imes 10^{-7}\ 	ext{m}^2$

Let's plug them in carefully:

$$K = \frac{0.2 	imes 4 	imes 10^{-7}}{8 	imes 10^{-7}}$$

The $10^{-7}$ terms cancel out immediately, simplifying our calculation:

$$K = \frac{0.8}{8} = 0.1	ext{ V/m}$$

So, the potential gradient of the potentiometer wire is exactly **$0.1	ext{ V/m}$**. This perfectly matches option (d). A straightforward and elegant application of fundamental concepts!

Select Answer:

Visualized Solution

$P o t e n t i o m e t er S e t u p$

$P o t e n t ia l G r a d i e n t (K)$

$O h m^{'} s L a w f or t h e W i r e$

$F or m u l a f or K$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Measuring Instruments

The Essence of a Potentiometer

What is Potential Gradient?

Connecting Ohm's Law

The Master Formula

Final Calculation

Similar Questions

The length of a wire of a potentiometer is $100 cm$ and the emf of its standard cell is $E$ volt. It is employed to measure the emf of a battery whose internal resistance is $0.5 Ω$ . If the balance point is obtained at $l = 30 cm$ from the positive end, the emf of the battery is

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is $1 k Ω$ . How much was the resistance on the left slot before interchanging the resistances?

Select Answer:

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; } Potential Gradient (K)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm′s Law for the Wire

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

.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; } Calculation

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