In a potentiometer experiment,when three cell | vedclass

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(C) Let $x$ be the potential gradient of the potentiometer wire. The emf $E$ of a cell is proportional to the balancing length $l$,i.e.,$E = xl$.For c

Vedclass · Accepted Answer

$1:1.2:2$

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(C) Let $x$ be the potential gradient of the potentiometer wire. The emf $E$ of a cell is proportional to the balancing length $l$,i.e.,$E = xl$.For cells $A, B, C$ in series: $E_A + E_B + E_C = x(420) \quad (1)$For cells $A, B$ in series: $E_A + E_B = x(220) \quad (2)$For cells $B, C$ in series: $E_B + E_C = x(320) \quad (3)$Subtracting equation $(2)$ from $(1)$: $E_C = x(420 - 220) = x(200)$.Substituting $E_C$ into equation $(3)$: $E_B + x(200) = x(320) \implies E_B = x(120)$.Substituting $E_B$ into equation $(2)$: $E_A + x(120) = x(220) \implies E_A = x(100)$.Thus,the ratio $E_A : E_B : E_C = 100 : 120 : 200 = 1 : 1.2 : 2$.

Similar Questions

$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

The arrangement as shown in the figure is called:

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