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In the circuit shown in figure, a cell of emf $ε$ and internal resistance r is connected across a load of resistance R, i is current in the circuit. $\bar{E}$ Represents the electrostatic field. ${\bar{E}}^{*}$ Represents extraneous field inside the cell which is non electrostatic in nature, which is responsible for motion of positive ions form negative terminal to positive terminal inside cell. Then choose the CORRECT option (s) about the line integral along cyclic path ABCDA.

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$\oint (\bar{E} + {\bar{E}}^{*}) . \bar{d l} = ε$

$\oint E^{*} . \bar{d l} = ε$

$\oint \bar{E} . \bar{d l} = i R$

$\oint \bar{E} . \bar{d l} = 0$

answer is A, B, D.

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Detailed Solution

Across the terminals of cell:

$- \int_{1}^{2} \bar{E} . \bar{d l} = Δ V = V_{2} - V_{1}$

$- \int_{1}^{2} {\bar{E}}^{*} . \bar{d l} = ε : e m f o f c e l l$

$- \int_{1}^{2} {\bar{E}}_{n e t} . \bar{d l} = \int_{1}^{2} (\bar{E} + {\bar{E}}^{*}) . \bar{d l} = ε - Δ V$

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