Q7 of 26 Page 1

A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit.


OR


In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Let the current in circuit be i.


Applying Kirchhoff’s law in loop ABCDA


10 + 38i = 200


38i = 200-10 = 190



∴ the current in circuit is 5A.


OR


Given


The balance point of the cell in open circuit = l1 = 350 cm.


the balance point of the cell with resistance = l2 = 300cm.


External resistance R = 9Ω.


We know that internal resistance of a primary cell by a potentiometer is given as,





∴ the internal resistance of cell is 1.5Ω .


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