A B is a potentiometer wire of length 100 cm and its resistance is 10 ohms. It is connected in series with a resistance R=40 ohms and a battery of e.m.f. 2 V and negligible internal resistance. If a source of unknown e.m.f. E is balanced by 40 cm length of the potentiometer wire, the value of E is

Question

A B is a potentiometer wire of length 100 cm and its resistance is 10 ohms. It is connected in series with a resistance R=40 ohms and a battery of e.m.f. 2 V and negligible internal resistance. If a source of unknown e.m.f. E is balanced by 40 cm length of the potentiometer wire, the value of E is (A) 0.8 V (B) 1.6 V (C) 0.08 V (D) 0.16 V

Tardigrade Admin · Accepted Answer

E=(e/(R+Rh+r)) (R/L) × l =(2/(10+40+0)) × (10/1) × 0.4=0.16 V .

Q. AB is a potentiometer wire of length 100cm and its resistance is 10ohms. It is connected in series with a resistance R=40 ohms and a battery of e.m.f. 2V and negligible internal resistance. If a source of unknown e.m.f. E is balanced by 40cm length of the potentiometer wire, the value of E is

0.8 V

1.6 V

0.08 V

0.16 V

E=(R+Rh​+r)e​LR​×l =(10+40+0)2​×110​×0.4=0.16V.

$E = \frac{e}{( R + R _{h} + r )} \frac{R}{L} \times l$
$= \frac{2}{( 10 + 40 + 0 )} \times \frac{10}{1} \times 0.4 = 0.16 V .$