Question:medium

When cell of e.m.f. ' \( E_1 \) ' is connected to potentiometer wire, the balancing length is ' \( l_1 \) '. Another cell of e.m.f. ' \( E_2 \) ' (\( E_1>E_2 \)) is connected so that two cells oppose each other, the balancing length is ' \( l_2 \) '. The ratio \( E_1 : E_2 \) is

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In opposition method: $E_1 - E_2 = k l_{opp}$ and $E_1 + E_2 = k l_{sum}$. Here $E_1$ alone is compared to opposition.
Updated On: May 7, 2026
  • \( \frac{l_1}{l_1+l_2} \)
  • \( \frac{l_1}{l_1-l_2} \)
  • \( \frac{l_1-l_2}{l_1} \)
  • \( \frac{l_1+l_2}{l_1-l_2} \)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
A potentiometer measures the unknown electromotive force (e.m.f.) of a cell by finding a point on the potentiometer wire where the potential drop equals the cell's e.m.f.
The principle is that the potential drop across any length of a uniform wire is directly proportional to that length (\( E \propto l \)).
Step 2: Key Formula or Approach:
The basic potentiometer equation is: \[ E = k \cdot l \] where \( E \) is the e.m.f., \( l \) is the balancing length, and \( k \) is the potential gradient (voltage per unit length) of the wire.
When cells are connected in opposition, their net e.m.f. is the difference of their individual e.m.f.s.
Step 3: Detailed Explanation:
In the first case, only the cell with e.m.f. \( E_1 \) is connected. The balancing length is given as \( l_1 \).
Using the potentiometer principle: \[ E_1 = k l_1 \quad \dots \text{(Equation 1)} \] In the second case, another cell with e.m.f. \( E_2 \) is connected in opposition to \( E_1 \). Since \( E_1>E_2 \), the effective net e.m.f. is \( (E_1 - E_2) \).
The balancing length for this combination is given as \( l_2 \).
Applying the potentiometer principle again: \[ E_1 - E_2 = k l_2 \quad \dots \text{(Equation 2)} \] We need to find the ratio \( \frac{E_1}{E_2} \).
From Equation 2, we can isolate \( E_2 \): \[ E_2 = E_1 - k l_2 \] Substitute \( E_1 \) from Equation 1 into this expression: \[ E_2 = k l_1 - k l_2 \] \[ E_2 = k (l_1 - l_2) \] Now, calculate the required ratio: \[ \frac{E_1}{E_2} = \frac{k l_1}{k (l_1 - l_2)} \] The potential gradient \( k \) cancels out: \[ \frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2} \] Step 4: Final Answer:
The ratio \( E_1 : E_2 \) is \( \frac{l_1}{l_1-l_2} \).
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