In the potentiometer circuit shown in figure the internal resistance of the 6 V

Discussion Forum : Current Electricity

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)

Options:

A . 0.7 V

B . 0.8 V

C . 0.9 V

D . 1 V

Answer: Option C
:
C

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
$= \frac{3}{4} \times \frac{3 \times 60}{100} = 0.9 V$

Was this answer helpful ?

Next Question

Discussion Forum : Current Electricity

Submit Your Solution hear: