The potential difference across the ends of a | vedclass

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

Question

(B) Let the potential gradient of the potentiometer wire be $k$.Let the $EMF$ of the two cells be $E_1$ and $E_2$.When the cells assist each other,the

Vedclass · Accepted Answer

$3:1$

Vedclass · Answer

(B) Let the potential gradient of the potentiometer wire be $k$.Let the $EMF$ of the two cells be $E_1$ and $E_2$.When the cells assist each other,the resultant $EMF$ is $(E_1 + E_2)$.According to the principle of the potentiometer,$(E_1 + E_2) = k 	imes 120 \ cm$  ---$(i)$When the cells oppose each other,the resultant $EMF$ is $(E_1 - E_2)$.According to the principle of the potentiometer,$(E_1 - E_2) = k 	imes 60 \ cm$  ---(ii)Dividing equation $(i)$ by equation (ii):$\frac{E_1 + E_2}{E_1 - E_2} = \frac{120}{60} = \frac{2}{1}$Applying componendo and dividendo or cross-multiplying:$E_1 + E_2 = 2(E_1 - E_2)$$E_1 + E_2 = 2E_1 - 2E_2$$3E_2 = E_1$Therefore,the ratio $\frac{E_1}{E_2} = \frac{3}{1}$.

Similar Questions

$A$ potentiometer having a potential gradient of $2\, mV/cm$ is used to measure the potential difference across a resistance of $10\, \Omega$. If a length of $50\, cm$ of the potentiometer wire is required to obtain the null point,the current passing through the $10\, \Omega$ resistor is (in $mA$):

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

In a potentiometer experiment,a cell is balanced at a length of $240 \ cm$. When the cell is shunted with a resistance of $2 \ \Omega$,it is balanced at a length of $120 \ cm$. The internal resistance of the cell is .......... $\Omega$.

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In potentiometer experiments,two cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$,the balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $E_1 / E_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module