Question

AB ia a potentiate wire of length 100 cm and is resistance is $10\Omega$. It is connected in series with a resistance $R=40\Omega$ and a battery of emf 2 V and negligible internal resistance. If a source of unknown potentiate wire, the value of E is:

• A. 0.8 V
• B. 1.6 V
• C. 0.06 V
• D. 0.16 V

Answer 1

The correct option is D 0.16 V

Given,

Cell voltage, $V=2V$

$I=\frac{V}{R+r}=\frac{2}{40+10}=0.04A$

Potential in wire= $Ir=0.04\times 10=0.4V$

Potential gradient, $\frac{0.4}{100}=0.004Vc{m}^{-1}$

Emf of cell is equal to length 40 cm =$0.004\times 40=0.16V$

Hence, value of E is $0.16V$