In a potentiometer experiment, when three cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are connected in…

In a potentiometer experiment, when three cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are connected in series, the balancing length is found to be $420 \mathrm{~cm}$. If cells A and B are connected in series the balancing length is $220 \mathrm{~cm}$ and for cells B and C connected in series the balancing length is $320 \mathrm{~cm}$. The emf of cells A, B and C are respectively in the ratio of
  1. 2:3:5
  2. 5:4:3
  3. 1:1.2:2
  4. 1.2:1:2

Solution

$\begin{aligned} & \mathrm{E}_{\mathrm{A}}+\mathrm{E}_{\mathrm{B}}+\mathrm{E}_{\mathrm{C}}=\mathrm{x}(420) \quad[\mathrm{x}=\text { potential gradient }] \\ & \mathrm{E}_{\mathrm{A}}+\mathrm{E}_{\mathrm{B}}=\mathrm{x}(220) \quad \therefore \mathrm{E}_{\mathrm{C}}=\mathrm{x}(420)-\mathrm{x}(220)=\mathrm{x}(200) \\ & \mathrm{E}_{\mathrm{B}}+\mathrm{E}_{\mathrm{C}}=\mathrm{x}(320) \quad \therefore \mathrm{E}_{\mathrm{B}}=\mathrm{x}(320)-\mathrm{x}(200)=\mathrm{x}(120) \\ & \mathrm{E}_{\mathrm{A}}=\mathrm{x}(220)-\mathrm{E}_{\mathrm{B}}=\mathrm{x}(100) \\ & \therefore \mathrm{E}_{\mathrm{A}}: \mathrm{E}_{\mathrm{B}}: \mathrm{E}_{\mathrm{C}}=100: 120: 200=1: 1.2: 2\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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