AB is a potentiometer wire of length 100 cm and its resistance is 10 - It is connected in series with a battery of emf 2 V and resistance 40 - If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire- the value of E isA- 0-8 VB- 1-6 VC- 0-08 VD- 0-16 V

The correct option is D 0.16 VPotential gradient nbsp; across potentiometer wire is nbsp; dV/dl = E_0 R_AB/(R_AB +R)L = 2 × 10/(10 + 40)1 = 0.4 Vm^ 1 U ...

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Standard XII

Physics

Electric Bulbs

AB is a poten...

Question

$A B$ is a potentiometer wire of length $100 c m$ and its resistance is $10 Ω$ . It is connected in series with a battery of emf $2 V$ and resistance $40 Ω$ . If a source of unknown emf $E$ is balanced by $40 c m$ length of the potentiometer wire, the value of $E$ is

0.8 V

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1.6 V

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0.08 V

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0.16 V

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Solution

The correct option is D $0.16 V$
Potential gradient across potentiometer wire is

$\frac{d V}{d l} = \frac{E_{0} R_{A B}}{(R_{A B} + R) L}$
$= \frac{2 \times 10}{(10 + 40) 1}$
$= 0.4 V m^{- 1}$

Unknown emf is
$E = \frac{d V}{d x} \times l$ $(∵ l = 40 c m)$
$E = 0.4 \times 40 \times 10^{- 2} = 0.16 V$

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AB is a potentiometer wire of length 100 cm and its resistance is 10 Ω. It is connected in series with a battery of emf 2 V and resistance 40 Ω. If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, the value of E is

The correct option is D 0.16 VPotential gradient across potentiometer wire is dVdl=E0RAB(RAB+R)L =2×10(10+40)1 =0.4 Vm−1 Unknown emf is E=dVdx×l (∵l=40 cm) E=0.4×40×10−2=0.16 V

$A B$ is a potentiometer wire of length $100 c m$ and its resistance is $10 Ω$ . It is connected in series with a battery of emf $2 V$ and resistance $40 Ω$ . If a source of unknown emf $E$ is balanced by $40 c m$ length of the potentiometer wire, the value of $E$ is