Uses of Potentiometer | Comparison of emf of two cells using Potentiometer || Class 12, JEE & NEET

The measuring instrument called a potentiometer is essentially a voltage divider used for measuring electric potential (voltage); the component is an implementation of the same principle, hence its name. Potentiometers are commonly used to control electrical devices such as volume controls on audio equipment. We will study here about Comparison of emf of two cells using Potentiometer, Determination of internal resistance of cell, Calibration of voltmeter, Calibration of ammeter, Measurement of small thermo emf.

The Potentiometer is mainly used for,

PRECAUTIONS IN THE USE OF POTENTIOMETER

### Determination of unknown EMF or Potential Difference

[3] If unknown potential difference V is balanced for length $\ell$ than $V=x \ell=\left(\frac{E_{0}}{\ell_{0}}\right) l$

[4] If the length of potentiometer wire is changed from L to L’ then the new balancing length is $\ell^{\prime}=\left(\frac{L^{\prime}}{L}\right) \ell$ If length is increased L’ > L so $l^{\prime}>l$ and if length is decreased L’ < L so $l^{\prime}<l$ So change in balancing length $\Delta \ell=\left(\ell^{\prime}-\ell\right)$

[ 5] If the current flowing through resistance $\mathrm{R}$ is I then

$$\mathrm{V}=\mathrm{IR}=\mathrm{x} \ell=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \ell \quad \text { so } \quad \mathrm{I}=\frac{\mathrm{x} \ell}{\mathrm{R}}=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \frac{\ell}{\mathrm{R}}$$

For determination of current we use a coil of standard resistance.

Ex. The current flowing through the primary circuit is 2A and resistance per unit length is

0.2 $\Omega / m$. If the potential difference across 10 ohm coil is balanced at 2.5 m then find current flowing through the coil.

Sol. The potential gradient $x=\frac{I R}{L}=2 \times 0.2=0.4 V / m$

Unknown potential $V=x \ell=I^{\prime} R$ so $I^{\prime}=\frac{x \ell}{R}=\frac{0.4 \times 2.5}{10}=0.1 A$

### Comparison of emf of two cells using Potentiometer

[1] No standardisation is necessary.

[2] Let $E_{1}$ emf be balanced at length $\ell_{1}$ and $E_{2}$ emf be balanced at length $\ell_{2}$ then

$E_{1}=x \ell_{1}$ and $E_{2}=x \ell_{2}$ so $\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

[3] If two cells joined in series support each other then the balancing length is $\ell_{1}$ so $E_{1}+E_{2}=x \ell_{1}$

If two cells joined in series oppose each other then the balancing length is $\ell_{2}$ so $E_{1}-E_{2}=x \ell_{2}$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{\ell_{1}}{\ell_{2}}$ or $\frac{E_{1}}{E_{2}}=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}}$

Ex. A uniform potential gradient is established across a potentiometer wire. Two cells of emf $E_{1}$ and $E_{2}$ connected to support and oppose each other are balanced over $\ell_{1}$ = 6m and $\ell_{2}$ = 2m. Find E1/E2.

Sol. $E _{1}+ E _{2}= x \ell_{1}=6 x$ and $E _{1}- E _{2}=2 x$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{6}{2}$ or $\frac{E_{1}}{E_{2}}=\frac{2}{1}$

### Determination of Internal Resistance of Cell

[1] No standardization is necessary.

[2] Keeping $\mathrm{K}_{1}$ open the balancing length $\ell_{1}$ gives emf of cell so E = $x \ell_{1}$

[3] Keeping $K_{1}$ closed the balancing length $\ell_{2}$ for some resistance R gives potential difference so V = $x \ell_{2}$

Internal resistance $r=\frac{E-V}{V} \quad R=\frac{x \ell_{1}-x \ell_{2}}{x \ell_{2}}$

$R=\left(\frac{\ell_{1}-\ell_{2}}{\ell_{2}}\right) R$

### Calibration of Voltmeter

[1] The voltmeters do not given accurate measurements because they do not have infinite resistance.

[2] The error in measurement is found by comparision with readings of potentiometer.

[3] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[4] The unknown potential difference V’ is balanced at length $\ell_{1}$ then $V^{\prime}=x \ell_{1}=\frac{E_{0}}{l_{0}} \ell_{1}$

[5] If reading of voltmeter is V then error is V – V’ which can be positive, negative or zero.

[6] The calibration curve is obtained plotting voltmeter reading V on x axis and error on y axis.

### Calibration of Ammeter

[1] V = IR and if R = 1$\Omega$ then V = I so potential difference across 1$\Omega$ resistance is equal to current through the resistance.

[2] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[3] The current through R or 1$\Omega$ coil is measured by

ammeter () and calculated by potentiometer as $V^{\prime}=I^{\prime} \times 1 \Omega=x \ell_{1}=\left(\frac{E_{0}}{\ell_{0}}\right) \ell_{1}$

[4] The error $I-I^{\prime}$can be found and calibration curve is obtained by plotting ammeter reading  on x axis and error on y axis.

### Measurement of Small Thermo EMF

[1] A high resistance box is used in primary circuit to reduce primary current.

[2] If galvanometer shows no deflection for some $R_{1}$ then potential difference across $R_{1}$ is $V=E_{0}=I R_{1}$ or $|=E_{0} / R_{1}$

[3] If balancing length for small thermo emf E is $\ell$ then $E=x \ell=\frac{E_{0}}{R_{1}} \cdot \frac{R}{L} \ell$

### Precautions in use of Potentiometer

[1] All high potential terminals of primary and secondary circuit should be connected at same point of Potentiometer.

[2] The known emf of primary circuit must be greater than the unknown potential difference to be measured in secondary circuit.

[3] The balancing length is always measured from the point of higher potential.

[4] In state of zero deflection no current flows in galvanometer circuit but it flows in primary circuit.

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

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Treatment of benzene with CO/HCl in the presence of anhydrous $\mathrm{AlCl}_{3} / \mathrm{CuCl}$ followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with

$\mathrm{Br}_{2} / \mathrm{Na}_{2} \mathrm{CO}_{3}$, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with $\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}$, followed by $\mathrm{H}_{3} \mathrm{PO}_{4}$ treatment gives Z as the major product.

(There are two questions based on PARAGRAPH “X”, the question given below is one of them)

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An organic acid P $\left(\mathrm{C}_{1} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

(There are two questions based on PARAGRAPH “A”, the question given below is one of them)

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Paragraph for questions 1 to 3

Copper is the most nobel of the first row transition metals and occurs in small deposits in several

countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_{4}, 5 \mathrm{H}_{2} \mathrm{O}\right),$ atacamite $\left(\mathrm{Cu}_{2} \mathrm{Cl}(\mathrm{OH})_{3}\right),$ cuprite $\left(\mathrm{Cu}_{2} \mathrm{O}\right),$ copper glance (Cu.S) and malachite $\left(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\right) .$ However, $80 \%$ of the world copper production comes from the ore chalcopyrite (CuFeS_{2} ) . \text { The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.

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Paragraph for Questions 2 to 3

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

$\mathrm{M}(\mathrm{s}) | \mathrm{M}^{+}\left(\mathrm{aq} ; 0.05 \text { molar) } \| \mathrm{M}^{+}(\mathrm{aq} ; 1 \mathrm{molar}) | \mathrm{M}(\mathrm{s})\right.$

For the above electrolytic cell the magnitude of the cell potential $\left|\mathrm{E}_{\mathrm{cell}}\right|$ = 70 mV.

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Paragraph for Question 6 and 7

The electrochemical cell shown below is a concentration cell.

$\mathrm{M} | \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt,

$\mathrm{MX}_{2}$) | | $\mathbf{M}^{2+}$ (0.001 mol $\mathrm{dm}^{-3}$) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

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Surface Chemistry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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Monotonicity – JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Monotonicity – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Maxima and Minima – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

3D Geometry- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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Definite integration – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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Given that for each a $\in(0,1)$, $\lim _{\mathrm{h} \rightarrow 0^{+}} \int_{\mathrm{h}}^{1-\mathrm{h}} \mathrm{t}^{-\mathrm{a}}(1-\mathrm{t})^{\mathrm{a}-1}$ $\mathrm{dt}$ exists. Let this limit be g(a). In addition,

it is given that the function g(a) is differentiable on (0,1).

# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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Let $\mathrm{F}: \mathbb{U} \rightarrow \square$ be a thrice differentiable function. Suppose that $\mathrm{F}(1)=0, \mathrm{F}(3)=-4 \mathrm{F}^{\prime}(\mathrm{x})<$ 0 for all $\mathrm{x} \in(1 / 2,3) .$ Let $f(\mathrm{x})=\mathrm{xF}(\mathrm{x})$ for all $\mathrm{x} \in \mathbb{D}$.

# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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# Why the sky is blue?

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