A potentiometer wire of length 4 , m and resi | vedclass

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

Question

(B) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r = 5 \, \Omega + 992 \, \Omega + 3 \, \Omega = 1000 \, \Omega$.
The

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(B) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r = 5 \, \Omega + 992 \, \Omega + 3 \, \Omega = 1000 \, \Omega$.
The current flowing through the potentiometer wire is $I = \frac{E}{R_{total}} = \frac{4 \, V}{1000 \, \Omega} = 0.004 \, A$.
The potential drop across the entire $4 \, m$ wire is $V_{wire} = I 	imes R_{wire} = 0.004 \, A 	imes 5 \, \Omega = 0.02 \, V$.
The potential gradient (potential drop per unit length) is $k = \frac{V_{wire}}{L} = \frac{0.02 \, V}{4 \, m} = 0.005 \, V/m$.
The e.m.f. balanced by a length of $0.75 \, m$ is $E' = k 	imes l = 0.005 \, V/m 	imes 0.75 \, m = 0.00375 \, V$.
Converting to millivolts, $E' = 0.00375 	imes 1000 \, mV = 3.75 \, mV$.

$A$ potentiometer wire of length $4 \, m$ and resistance $5 \, \Omega$ is connected in series with a resistance of $992 \, \Omega$ and a cell of e.m.f. $4 \, V$ with internal resistance $3 \, \Omega$. The length of $0.75 \, m$ on the potentiometer wire balances the e.m.f. of (in $ \, mV$)

Similar Questions

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

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 .......... $\Omega$.

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

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