Cells and Potentiometer

Electricity and Magnetism Level 4

Consider a potentiometer circuit,Primary cell is ideal.The length of potentiometer wire is $1$ m and the resistance per unit length of potentiometer wire varies with length as $\lambda=2x$ ohm/m,where $x$ is the distance from end $A$ .Resistance of Rheostat varies with time as $R=t^2$ ohm.Null deflection point for secondary cell is obtained at $x=\frac{1}{2}$ m and at $t=1$ sec.If emf of secondary cell is $\frac{1}{\gamma}$ times the emf of primary cell,find $\gamma$ .

The answer is 8.00.

1 solution

Rajdeep Brahma
Jun 27, 2018

$\lambda=2x$ . $R_{AB}=$ $\int_0^1 \lambda \, dx$ , $R_{AB}=1ohm$ , $R_{AP}=$ $\int_0^x \lambda \, dx$ = $\frac{x^2}{2}$ , $I=\frac{E_{P}}{R_{AB}+R}=\frac{E_{P}}{1+t^2}$ .Now $V_{AP}=IR_{AP}=E_{S}=\frac{E_{P}*x^2}{1+t^2}$ .At $t=1$ sec, $x=\frac{1}{2}$ . $E_{S}=\frac{E_{P}/4}{1+1}$ = $\frac{E_{P}}{8}$ . $\gamma=8.00$

Awesome question. Can you plz plz plz post more questions on Current electricity. Plz!

Md Zuhair - 2 years, 11 months ago

Oh liked it??well i will post

rajdeep brahma - 2 years, 11 months ago

Hey there's an typo in your solution. After R(AJ).. You have written x^2/2. But it's x^2

Md Zuhair - 2 years, 11 months ago

Cells and Potentiometer

The answer is 8.00.

1 solution

0 pending reports