A battery of emf E_0 = 12, V is connected acr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equivalent $emf$ $(E_{eq})$ of the two cells connected in parallel is given by the formula:$E_{eq} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r

Vedclass · Accepted Answer

$25\,cm$

Vedclass · Answer

(C) The equivalent $emf$ $(E_{eq})$ of the two cells connected in parallel is given by the formula:$E_{eq} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$Given $\varepsilon_1 = 2\,V, r_1 = 2\,\Omega$ and $\varepsilon_2 = 4\,V, r_2 = 6\,\Omega$:$E_{eq} = \frac{2 	imes 6 + 4 	imes 2}{2 + 6} = \frac{12 + 8}{8} = \frac{20}{8} = 2.5\,V$The total resistance of the potentiometer wire is $R_{wire} = 4\,m 	imes 4\,\Omega/m = 16\,\Omega$.The total resistance of the primary circuit is $R_{total} = R + R_{wire} = 8\,\Omega + 16\,\Omega = 24\,\Omega$.The current in the primary circuit is $I = \frac{E_0}{R_{total}} = \frac{12\,V}{24\,\Omega} = 0.5\,A$.The potential drop per unit length (potential gradient) is $x = I 	imes (resistance\,per\,unit\,length) = 0.5\,A 	imes 4\,\Omega/m = 2\,V/m$.At the balancing point $N$,the potential drop across length $l$ is equal to $E_{eq}$:$x 	imes l = E_{eq}$$2\,V/m 	imes l = 2.5\,V$$l = \frac{2.5}{2} = 1.25\,m = 125\,cm$.*Correction*: Re-evaluating the parallel combination based on the diagram polarity: $\varepsilon_1$ and $\varepsilon_2$ are opposing. $E_{eq} = \frac{\varepsilon_1/r_1 - \varepsilon_2/r_2}{1/r_1 + 1/r_2} = \frac{2/2 - 4/6}{1/2 + 1/6} = \frac{1 - 2/3}{4/6} = \frac{1/3}{2/3} = 0.5\,V$.Now,$2\,V/m 	imes l = 0.5\,V \implies l = 0.25\,m = 25\,cm$.

Similar Questions

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

In a potentiometer arrangement,$E_1$ is the cell establishing the current in the primary circuit,$E_2$ is the cell to be measured,$AB$ is the potentiometer wire,and $G$ is a galvanometer. Which of the following are the essential conditions for a balance point to be obtained?

$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module