A potentiometer wire of length 10 m and resistance 30 - is connected in series with a battery of emf 2-5 V- internal resistance 5 - and an external resistance R- If the fall of potential along the potentiometer wire is 50 - V mm-1- then the value of R is found to be 23 n - What is n-

If I be the current through the circuit.thus, #160;I(R+R_p+r)=E ⇒ I=E/R+R_p+r #160;where #160; E= emf of the circuit, R_p= resistance ...

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Standard XII

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire of length $10 m$ and resistance $30 Ω$ is connected in series with a battery of emf $2.5 V$ , internal resistance $5 Ω$ , and an external resistance $R$ . If the fall of potential along the potentiometer wire is $50 μ V m m^{- 1}$ , then the value of $R$ is found to be $23 n Ω$ . What is $n$ ?

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Solution

If $I$ be the current through the circuit.
thus, $I (R + R_{p} + r) = E \Rightarrow I = \frac{E}{R + R_{p} + r}$ where

$E =$ emf of the circuit, $R_{p} =$ resistance of wire, $r =$ internal resistance of cell.
Potential gradient , $k = \frac{I R_{p}}{L} = \frac{E R_{p}}{(R + R_{p} + r) L}$
$\frac{50 \times 10^{- 6}}{10^{- 3}} = \frac{2.5 (30)}{(R + 30 + 5) 10}$
$R + 35 = \frac{7.5}{50 \times 10^{- 3}} = 150 \Rightarrow R = 150 - 35 = 115$
thus, $23 n = 115 \Rightarrow n = \frac{115}{23} = 5$

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A potentiometer wire of length 10 m and resistance 30 Ω is connected in series with a battery of emf 2.5 V, internal resistance 5Ω, and an external resistance R. If the fall of potential along the potentiometer wire is 50 μVmm−1, then the value of R is found to be 23n Ω. What is n?

If I be the current through the circuit.thus, I(R+Rp+r)=E⇒I=ER+Rp+r where E=emf of the circuit, Rp= resistance of wire, r= internal resistance of cell.Potential gradient , k=IRpL=ERp(R+Rp+r)L50×10−610−3=2.5(30)(R+30+5)10R+35=7.550×10−3=150⇒R=150−35=115thus, 23n=115⇒n=11523=5