In the primary circuit of a potentiometer, a cell of E.M.F 1 V and a rheostat of `15 Omega` are connected in series. If the resistance of the potentiometer wire is `10 Omega` the minimum voltage at the ends of the wire (in V) will be

Options

A. `0.1`
B. `0.4`
C. `0.06`
D. `1.0`

Answer

To solve the problem step by step, we will use the principles of electrical circuits, specifically applying Kirchhoff's voltage law.

### Step 1: Understand the Circuit Configuration
We have a circuit consisting of:
- A cell with an EMF (E) of 1 V.
- A rheostat (variable resistor) with a resistance (R_H) of 15 Ω.
- A potentiometer wire with a resistance (R_W) of 10 Ω.

### Step 2: Apply Kirchhoff's Voltage Law
According to Kirchhoff's voltage law, the sum of the potential differences (voltage) in a closed loop is equal to the EMF of the sources in that loop.

The equation can be set up as follows:
\[ E - I \cdot R_W - I \cdot R_H = 0 \]

Where:
- \( E = 1 \, \text{V} \) (EMF of the cell)
- \( R_W = 10 \, \Omega \) (resistance of the potentiometer wire)
- \( R_H = 15 \, \Omega \) (resistance of the rheostat)
- \( I \) is the current flowing through the circuit.

### Step 3: Rearrange the Equation
Rearranging the equation gives:
\[ E = I \cdot (R_W + R_H) \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 1 = I \cdot (10 + 15) \]
\[ 1 = I \cdot 25 \]

### Step 5: Solve for Current (I)
Now, solve for the current \( I \):
\[ I = \frac{1}{25} \, \text{A} \]

### Step 6: Calculate the Minimum Voltage Across the Potentiometer Wire
The voltage across the potentiometer wire (V_AB) can be calculated using Ohm's Law:
\[ V_{AB} = I \cdot R_W \]
Substituting the values:
\[ V_{AB} = \left(\frac{1}{25}\right) \cdot 10 \]
\[ V_{AB} = \frac{10}{25} = 0.4 \, \text{V} \]

### Conclusion
The minimum voltage at the ends of the potentiometer wire is **0.4 V**.

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