A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 $Ω$ as shown in the figure. Find the value of current in the circuit.
OR
In a potentiometer arrangement for deter-mining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Q is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

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Detailed Solution

$Given : E_{1} = 10 V; E_{2} = 200 V, r = 38 Ω,$
$I = ?$
$Netemf, E = E_{2} - E_{1} (∵ twocells$
being in position
$= 200 V - 10 V = 190 V$
$∴ Current, I = \frac{Net emf}{Resistance} = \frac{190 V}{38 Ω} = 5 A .$

OR

Balance point in open circuit, $l_{1} = 350 cm$
$R = 9 Ω$
New balance point $l_{2} = 300 cm$
Internal resistance of a cell is given by,
$r = R (\frac{l_{1} - l_{2}}{l_{2}}) = 9 (\frac{350 - 300}{300}) = 1 .5 Ω$