A potentiometer wire of Length `L` and a resistance `r` are connected in series with a battery of e.m.f. `E_(0)` and a resistance `r_(1)`. An unknown e.m.f. `E` is balanced at a length `l` of the potentiometer wire. The e.m.f. `E` will be given by :

Options

A. `(LE_(0)r)/(lir_(1))`
B. `(E_(0)r)/((r+r_(1))).(l)/(L)`
C. `(E_(0)l)/(L)`
D. `(LE_(0)r)/((r+r_(1))l)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Circuit
We have a potentiometer wire of length \( L \) and resistance \( r \) connected in series with a battery of e.m.f. \( E_0 \) and a resistance \( r_1 \). An unknown e.m.f. \( E \) is balanced at a length \( l \) of the potentiometer wire.

### Step 2: Find the Current in the Circuit
The total resistance in the circuit is the sum of the resistance of the potentiometer wire and the external resistance:
\[ R_{\text{total}} = r + r_1 \]
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{E_0}{R_{\text{total}}} = \frac{E_0}{r + r_1} \]

### Step 3: Calculate the Potential Difference Across the Potentiometer Wire
The potential difference \( V_{AB} \) across the potentiometer wire can be calculated using:
\[ V_{AB} = I \cdot r \]
Substituting the expression for \( I \):
\[ V_{AB} = \left(\frac{E_0}{r + r_1}\right) \cdot r \]

### Step 4: Determine the Potential Gradient
The potential gradient \( K \) along the potentiometer wire is given by:
\[ K = \frac{V_{AB}}{L} \]
Substituting the expression for \( V_{AB} \):
\[ K = \frac{\left(\frac{E_0 \cdot r}{r + r_1}\right)}{L} = \frac{E_0 \cdot r}{L \cdot (r + r_1)} \]

### Step 5: Calculate the Unknown E.m.f. \( E \)
The e.m.f. \( E \) that corresponds to the balanced length \( l \) can be expressed as:
\[ E = K \cdot l \]
Substituting the expression for \( K \):
\[ E = \left(\frac{E_0 \cdot r}{L \cdot (r + r_1)}\right) \cdot l \]

### Final Expression
Thus, the unknown e.m.f. \( E \) is given by:
\[ E = \frac{E_0 \cdot r \cdot l}{L \cdot (r + r_1)} \]