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Byju's Answer

Standard XII

Physics

Potentiometer

In the potent...

Question

In the potentiometer circuit shown in figure the internal resistance of the 6 V battery is 1 ohm and the length of the wire AB is 100 cm. When AD = 60 cm the galvanometer shows no deflection. The emf of cell C is (the resistance of wire AB is 2 ohm)

0.7 V

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0.8 V

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0.9 V

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1 V

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Solution

The correct option is C 0.9 V
Current in the circuit due to 6 V battery is
$l = \frac{6}{1 + 5 + 2} = \frac{3}{4} A$
Now emf of cell C =potential difference across AD
$= \frac{3}{4} \times \frac{3 \times 60}{100} = 0.9 V$

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In the potentiometer circuit shown in figure the internal resistance of the 6 V battery is 1 ohm and the length of the wire AB is 100 cm. When AD = 60 cm the galvanometer shows no deflection. The emf of cell C is (the resistance of wire AB is 2 ohm)

The correct option is C 0.9 V Current in the circuit due to 6 V battery is l=61+5+2=34A Now emf of cell C =potential difference across AD =34× 3 × 60100=0.9 V

The correct option is C 0.9 V
Current in the circuit due to 6 V battery is
$l = \frac{6}{1 + 5 + 2} = \frac{3}{4} A$
Now emf of cell C =potential difference across AD
$= \frac{3}{4} \times \frac{3 \times 60}{100} = 0.9 V$