First slide

potentiometer

Question

AB and CD are two uniform resistance wires of lengths 100 cm and 80 cm, respectively. The connections are shown in the figure. The cell of emf 5 V is ideal while the other cell of emf E has internal resistance $2 Ω$ A length of 20 cm of wire CD is balanced by 40 cm of wire AB. Find the emf E in volt, if the reading of the ideal ammeter is 2 A. The other connecting wires have negligible resistance.

Difficult

A

10
B

12
C

8
D

zero

Solution

Potential difference across wire AB = 5 V
Potential difference across 40 cm of this wire $\frac{5}{100} \times 40 = 2 V .$
So potential difference across 20 cm of wire CD = 2V.
$∴$ Potential difference across wire CD $= \frac{2}{20} \times 80 = 8 V$
Potential difference across $2 Ω$ resistor = 2 x 2 = 4 V
$∴$ Emf of the cell = 8 + 4 = 12 V

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