Question # potential gradient in terms of resistivity of the potentiometer wire. Figure shows a long potentiometer wire ( A B ) having a constant potential gradient. The null points for the two primary cells of emfs ( mathrm{E}_{1} ) and ( mathrm{E}_{2} ) connected in the manner shown are obtained at a distance of ( 11=120 mathrm{cm} ) and ( 12=300 mathrm{cm} ) from the end ( mathrm{A} )

# potential gradient in terms of resistivity of the potentiometer wire. Figure shows a long potentiometer wire ( A B ) having a constant potential gradient. The null points for the two primary cells of emfs ( mathrm{E}_{1} ) and ( mathrm{E}_{2} ) connected in the manner shown are obtained at a distance of ( 11=120 mathrm{cm} ) and ( 12=300 mathrm{cm} ) from the end ( mathrm{A} )

Determine: ( E_{1} / E_{2} ) and (ii) position of null point for the cell ( E_{1} ) only.

Solution

[

begin{array}{l}

varepsilon_{1}-varepsilon_{2}=k(120)

varepsilon_{1}+varepsilon_{2}=K(300)

end{array}

]

( Rightarrow frac{varepsilon_{1}-varepsilon_{2}}{varepsilon_{1}+varepsilon_{2}}=frac{120}{300} )

( Rightarrow frac{2 varepsilon_{1}}{-2 varepsilon_{2}}=frac{420}{-180} )

( Rightarrow mid frac{epsilon_{1}}{varepsilon_{2}}=frac{7}{3} )

( Rightarrow quad varepsilon_{2}=frac{3 varepsilon_{1}}{7} )

( epsilon_{1}-frac{3 varepsilon_{1}}{7}=k(120) )

( Rightarrow quad frac{4 G}{7}=k(120) )

( Leftrightarrow quad G=k(210) )

( Rightarrow ) Null pout for ( varepsilon_{1} ) cell oul ( y=210 mathrm{cm} )