In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between

Q: Why do we add the internal resistance of the cells to the total resistance?

The cells are in series with the main driving circuit. Their internal resistance acts just like any other series resistor, restricting the total current flowing through the potentiometer wire.

Q: Does the voltmeter draw any current from the circuit?

No. The problem explicitly states it's an 'ideal voltmeter', which means it has infinite resistance. Therefore, it draws zero current and doesn't affect the main circuit's current.

Q: Why is the length 400 cm when the diagram shows 4 separate wires?

Each wire segment is 100 cm long, and they are connected in series by thick metal strips at the ends. Electrically, it behaves as a single continuous 400 cm wire.

Q: What if the voltmeter wasn't ideal?

If it wasn't ideal, it would draw some current. This would create a parallel circuit across the 50 cm segment, changing its effective resistance and altering the main current. The calculation would be much more complex.

Q: Why are the two cells' EMFs added together?

They are connected in series with the correct polarity (positive to negative) to drive current in the same direction, so their voltages add up to 3 V.

Question

In the circuit shown, a four-wire potentiometer is made of a $400	ext{ cm}$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r = 0.01\,\Omega/	ext{cm}$. If an ideal voltmeter is connected as shown with jockey $J$ at $50	ext{ cm}$ from end $A$, the expected reading of the voltmeter will be

Options:

(A) 0.20 V

(B) 0.75 V

(C) 0.25 V

(D) 0.50 V

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Welcome to this interesting problem on potentiometers. Look at the circuit carefully. We have a four-wire potentiometer, where each segment is $100	ext{ cm}$ long, making the total length $L = 400	ext{ cm}$. The resistance per unit length is given as $\lambda = 0.01\,\Omega/	ext{cm}$.

First, let's find the total resistance of this potentiometer wire from $A$ to $B$. Since we know the length and the resistance per centimeter, we just multiply them:

$$R_{AB} = \lambda 	imes L = 0.01 	imes 400 = 4\,\Omega$$

### The Master Equation
Now, let's look at the driving circuit on the left. We have two cells connected in series, each of $1.5	ext{ V}$, giving a total EMF of $3	ext{ V}$. Their internal resistances add up to $1\,\Omega$. Plus, there's an external resistor of $1\,\Omega$.

With all resistances known, we can find the steady current flowing through the main circuit. The total resistance is $4 + 1 + 1$, which is $6\,\Omega$. Dividing the net EMF of $3	ext{ V}$ by $6\,\Omega$, we get:

$$I = \frac{E_{	ext{net}}}{R_{AB} + r_{	ext{net}} + R_{	ext{ext}}} = \frac{3}{4 + 1 + 1} = 0.5	ext{ A}$$

### Final Calculation
Next, focus on the voltmeter. It's connected across a $50	ext{ cm}$ segment of the wire starting from $A$. What is the resistance of just this part? We multiply $0.01$ by $50$, which gives us $0.5\,\Omega$.

$$R_{50} = 0.01 	imes 50 = 0.5\,\Omega$$

Finally, the voltmeter reads the potential drop across this $50	ext{ cm}$ segment. Using Ohm's law, $V = I 	imes R$. Multiplying our current of $0.5	ext{ A}$ by the resistance of $0.5\,\Omega$, we get exactly $0.25	ext{ V}$.

$$V = I 	imes R_{50} = 0.5 	imes 0.5 = 0.25	ext{ V}$$

This problem beautifully illustrates the principle of a potentiometer. As long as the main current is constant, the potential drop is directly proportional to the balancing length.

Select Answer:

Visualized Solution

Circuit Analysis

Resistance of Wire $A B$

Driving Circuit Parameters

Main Circuit Current

Resistance of $50 cm$ Segment

Voltmeter Reading

Conceptual Takeaway

The Sigma Insight: Electrical Instruments

Analyzing the Setup Welcome to this interesting problem on potentiometers

The Master Equation Now, let's look at the driving circuit on the left

Final Calculation Next, focus on the voltmeter

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