A potentiometer wire of length 10 m and a resistance 30 - is connected in series with a battery of emf 2-5 V and internal resistance 5 - and an external resistance R- If the fall of potential along the potentiometer wire is 50 - V-mm- the value of R is in -

By using x=e/(R+R_h+r).R/L ⇒50 × 10^ 6/10^ 3 = 2.5/(30+R+5)×30/10⇒ R = 115 ...

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Byju's Answer

Standard XII

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire of length 10 m and a resistance $30 Ω$ is connected in series with a battery of emf 2.5 V and internal resistance $5 Ω$ and an external resistance R. If the fall of potential along the potentiometer wire is $50 μ V / m m$ , the value of R is (in $Ω$ )

115

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100

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Solution

The correct option is A 115
By using $x = \frac{e}{(R + R_{h} + r)} . \frac{R}{L}$
$\Rightarrow \frac{50 \times 10^{- 6}}{10^{- 3}} = \frac{2.5}{(30 + R + 5)} \times \frac{30}{10} \Rightarrow R = 115$

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A potentiometer wire of length 10 m and a resistance 30Ω is connected in series with a battery of emf 2.5 V and internal resistance 5Ω and an external resistance R. If the fall of potential along the potentiometer wire is 50 μV/mm, the value of R is (in Ω)

The correct option is A 115 By using x=e(R+Rh+r).RL ⇒50×10−610−3=2.5(30+R+5)×3010⇒R=115

A potentiometer wire of length 10 m and a resistance $30 Ω$ is connected in series with a battery of emf 2.5 V and internal resistance $5 Ω$ and an external resistance R. If the fall of potential along the potentiometer wire is $50 μ V / m m$ , the value of R is (in $Ω$ )

The correct option is A 115
By using $x = \frac{e}{(R + R_{h} + r)} . \frac{R}{L}$
$\Rightarrow \frac{50 \times 10^{- 6}}{10^{- 3}} = \frac{2.5}{(30 + R + 5)} \times \frac{30}{10} \Rightarrow R = 115$