A potentiometer wire of length $$L$$ and a resistance $$r$$ are connected in series with a battery of EMF $$E_{0 }$$ and resistance $$r_{1}$$. An unknown EMF is balanced at a length l of the potentiometer wire. The EMF $$E$$ will be given by:
1. $$\frac{L E_{0} r}{l r_{1}}$$
2. $$\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}$$
3. $$\frac{E_{0} l}{L}$$
4. $$\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}$$

Subtopic:  Meter Bridge & Potentiometer |
76%
From NCERT
NEET - 2015
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A potentiometer wire has a length of $$4​​~\text{m}$$ and resistance  $$8~\Omega.$$  The resistance that must be connected in series with the wire and an energy source of emf $$2~\text{V}$$, so as to get a potential gradient of $$1~\text{mV}$$ per cm on the wire is:
1. $$32~\Omega$$
2. $$40~\Omega$$
3. $$44~\Omega$$
4. $$48~\Omega$$

Subtopic:  Meter Bridge & Potentiometer |
65%
From NCERT
NEET - 2015
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The figure given below shows a circuit when resistances in the two arms of the meter bridge are $$5~\Omega$$ and $$R$$, respectively. When the resistance $$R$$ is shunted with equal resistance, the new balance point is at $$1.6l_1$$. The resistance $$R$$ is:

1. $$10~\Omega$$
2. $$15~\Omega$$
3. $$20~\Omega$$
4. $$25~\Omega$$

Subtopic:  Meter Bridge & Potentiometer |
73%
From NCERT
AIPMT - 2014
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A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $$2.0~\text{V}$$ and negligible internal resistance. The potentiometer wire itself is $$4~\text{m}$$ long. When the resistance, $$R$$, connected across the given cell, has values of (i) infinity (ii) $$9.5$$, the 'balancing lengths, on the potentiometer wire, are found to be $$3~\text{m}$$ and $$2.85~\text{m}$$, respectively. The value of the internal resistance of the cell is (in ohm):
1. $$0.25$$
2. $$0.95$$
3. $$0.5$$
4. $$0.75$$

Subtopic:  Meter Bridge & Potentiometer |
65%
From NCERT
AIPMT - 2014
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A potentiometer circuit is set up as shown in the figure below. The potential gradient across the potentiometer wire is k volt/cm. Ammeter present in the circuit reads 1.0 A when the two-way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths ${l}_{1}$ $cm$ and ${l}_{2}$ $cm$ respectively. The magnitudes of the resistors R and X in ohm, are then, respectively, equal to:

1. $k\left({l}_{2}-{l}_{1}\right)$ $and$ $k{l}_{2}$

2. $k{l}_{1}$ $and$ $k\left({l}_{2}-{l}_{1}\right)$

3. $k\left({l}_{2}-{l}_{1}\right)$ $and$ $k{l}_{1}$

4. $k{l}_{1}$ $and$ $k{l}_{2}$

Subtopic:  Meter Bridge & Potentiometer |
65%
From NCERT
AIPMT - 2010
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A cell can be balanced against 100 cm and 110 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is:

1.  1.0 Ω

2.  0.5 Ω

3.  2.0 Ω

4.  zero

Subtopic:  Meter Bridge & Potentiometer |
76%
From NCERT
AIPMT - 2008
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