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Byju's Answer

Standard XII

Physics

EMF and EMF Devices

A potentiomet...

Question

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_{0}$ and a resistance $r_{1}$ . An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by.

\frac{E_{0} r}{(r + r_{1})} \cdot \frac{l}{L}

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\frac{E_{0} l}{L}

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\frac{L E_{0} r}{(r + r_{1}) l}

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\frac{L E_{0} r}{l r_{1}}

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Solution

The correct option is A $\frac{E_{0} r}{(r + r_{1})} \cdot \frac{l}{L}$
Current in ABCD
$I = \frac{V}{R_{e q}} = \frac{V}{r + r_{1}}$
potential difference across potentiometr $= I r = \frac{V r}{(r + r_{1})}$
Potential gradient $= \frac{V}{L} = \frac{V r}{(r + r_{1})} \times \frac{1}{L} = k$
we know that EMF=kl
l=balancing length
$E M F = (\frac{V r}{r + r_{1}}) \frac{l}{L} = (\frac{ϵ r}{r + r_{1}}) \frac{l}{L}$

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A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by.

The correct option is A E0r(r+r1)⋅lLCurrent in ABCDI=VReq=Vr+r1potential difference across potentiometr =Ir=Vr(r+r1)Potential gradient =VL=Vr(r+r1)×1L=kwe know that EMF=kll=balancing lengthEMF=(Vrr+r1)lL=(ϵrr+r1)lL

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_{0}$ and a resistance $r_{1}$ . An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by.