Current Electricity 4 Question 10

10. A potentiometer wire AB having length L and resistance 12r is joined to a cell D of EMF ε and internal resistance r. A cell C having emf ε2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is

(2019 Main, 10 Jan I)

(a) 512L

(b) 1112L

(c) 1324L

(d) 1124L

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Solution:

  1. Given, length of potentiometer wire (AB)=L

Resistance of potentiometer wire (AB)=12r

EMF of cell D of potentiometer =ε

Internal resistance of cell ’ D=r

EMF of cell ’ C=ε2

Internal resistance of cell ’ C=3r

Current in potentiometer wire

i= EMF of cell of potentiometer  Total resistance of potentiometer circuit i=εr+12r=ε13r

Potential drop across the balance length AJ of potentiometer wire is

VAJ=i×RAJVAJ=i (Resistance per unit length of  potentiometer wire × length AJ)VAJ=i12rL×x

where, x =balance length AJ.

As null point occurs at J so potential drop across balance length AJ= EMF of the cell ’ C ‘.

VAJ=ε2i12rL×x=ε2

ε13r×12rL×x=ε2x=1324L



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