A potentiometer wire of length 1 m and resistance 10- is connected in series with a cell of e-m-f 2 volt with internal resistance 1 - and a resistance box including a resistance R- If the potential difference between the ends of the wire is 1 mV- then the value of R will be

I=V_2/R_1=1× 10^ 3/10=10^ 4A E=i[R+r+R_L] 2=10^ 4[R+1+10] 20000=R+11 R=20000 11=19989Ω ...

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Standard XI

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire of length 1 m and resistance $10 Ω$ is connected in series with a cell of e.m.f 2 volt with internal resistance $1 Ω$ and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

9989 Ω

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10000 Ω

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19989 Ω

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20000 Ω

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Solution

The correct option is C $19989 Ω$
$i = \frac{V_{2}}{R_{1}} = \frac{1 \times 10^{- 3}}{10} = 10^{- 4} A E = i [R + r + R_{L}] 2 = 10^{- 4} [R + 1 + 10] 20000 = R + 11 R = 20000 - 11 = 19989 Ω$

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A potentiometer wire of length 1 m and resistance 10Ω is connected in series with a cell of e.m.f 2 volt with internal resistance 1Ω and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

The correct option is C 19989Ωi=V2R1=1×10−310=10−4AE=i[R+r+RL]2=10−4[R+1+10]20000=R+11R=20000−11=19989Ω

A potentiometer wire of length 1 m and resistance $10 Ω$ is connected in series with a cell of e.m.f 2 volt with internal resistance $1 Ω$ and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

The correct option is C $19989 Ω$
$i = \frac{V_{2}}{R_{1}} = \frac{1 \times 10^{- 3}}{10} = 10^{- 4} A E = i [R + r + R_{L}] 2 = 10^{- 4} [R + 1 + 10] 20000 = R + 11 R = 20000 - 11 = 19989 Ω$