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 follow these instructions:

### Step 1: Understand the Circuit
We have a potentiometer circuit with:
- A cell of EMF (E) = 1 V
- A rheostat (R_r) = 15 Ω
- A potentiometer wire with resistance (R_w) = 10 Ω

### Step 2: Identify the Total Resistance in the Primary Circuit
The total resistance (R_total) in the primary circuit is the sum of the resistance of the rheostat and the potentiometer wire:
\[ R_{total} = R_r + R_w = 15 \, \Omega + 10 \, \Omega = 25 \, \Omega \]

### Step 3: Calculate the Current in the Circuit
Using Ohm's Law, we can find the current (I) flowing through the circuit. The formula is:
\[ I = \frac{E}{R_{total}} \]
Substituting the values:
\[ I = \frac{1 \, V}{25 \, \Omega} = 0.04 \, A \]

### Step 4: Calculate the Voltage Across the Potentiometer Wire
The voltage (V_AB) across the potentiometer wire can be calculated using Ohm's Law:
\[ V_{AB} = I \times R_w \]
Substituting the values:
\[ V_{AB} = 0.04 \, A \times 10 \, \Omega = 0.4 \, V \]

### Step 5: Conclusion
The minimum voltage at the ends of the potentiometer wire is:
\[ V_{AB} = 0.4 \, V \]

### Final Answer
The minimum voltage at the ends of the wire is **0.4 V**.

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