Ammeter and Voltmeter Circuit Diagram | Current Electricity Class 12, JEE & NEET
Voltmeters and ammeters are used to measure voltage and current, respectively. Here we will discuss both with Ammeter and Voltmeter Circuit Diagram.

## Ammeter

1. An ammeter is a low resistance galvanometer used to measure strength of current in an electrical circuit.
2. An ammeter is always connected in series in a circuit because when an ammeter is connected in series it does not appreciably change the resistance of circuit and hence the main current flowing through the circuit.
3. In ideal ammeter has zero resistance.
4. The reading of an ammeter is always less than actual current in the circuit because all practical ammeters have low finite resistance.
5. Smaller is the resistance of an ammeter more accurate will be the reading.
6. A galvanometer can be converted to an ammeter by connecting a low resistance shunt in parallel to coil of galvanometer. Ammeter Circuit Diagram: Here $\mathrm{I}_{9} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ or $\quad S=\frac{I_{g}}{I-I_{g}} G$ Here G is resistance of galvanometer and $I_{g}$ is current required to produced full scale deflection of current.
7. Shunt (S) : It is a low resistance connected in parallel to coil of galvanometer to convert it to ammeter. It protects a galvanometer from strong currents. It is also used to alter range of an ammeter.
8. The resistance of converted ammeter is $R_{A}=\frac{G S}{G+S}$
9. The range of an ammeter is increased by reducing shunt resistance S. If $\mathrm{I}=\mathrm{N} \mathrm{I}_{\mathrm{g}}$ then $S=\frac{I_{g}}{N I_{g}-I_{g}} G=\frac{G}{N-1}$ The range of an ammeter can be increased N times by reducing shunt to S = G/N-1.
10. It is possible to increase the range of an ammeter because it lowers the resistance of ammeter further.
11. The length of the shunt required $\ell=\pi r^{2} S / \rho$ where r is radius of shunt wire and  is specific resistance of material of shunt wire.
12. Reducing the shunt resistance may increase range but it reduces the sensitivity.

## Voltmeter

1. A voltmeter is a high resistance galvanometer used to measure potential difference.
2. A voltmeter is connected in parallel to a circuit element because when connected in parallel it draws least current from the main current. So it measures nearly accurate potential difference.
3. An ideal voltmeter has infinite resistance.
4. The reading of a voltmeter is always less than actual value because all practical voltmeter may have large but finite resistance.
5. Greater is the resistance of voltmeter more accurate is its reading.
6. A galvanometer is converted to a voltmeter by connecting a high resistance in series with coil of galvanometer. $V=I_{g}(R+G) \quad$ or $\quad R=\frac{V}{I_{g}}-G$
7. The resistance of converted voltmeter is $\mathrm{R}_{\mathrm{v}}=\mathrm{R}+\mathrm{G}$
8. The range of a voltmeter is increased by increasing the series resistance. If $\quad \mathrm{V}=\mathrm{NV}_{9}=\mathrm{NI}_{9} \mathrm{G} \quad$ then $\quad R=\frac{N I_{9} G-G}{I_{9}}=(N-1) G$ The value of resistance required to increase range $N$ times is $R=(N-1) G$

Ex. What is the value of shunt which passes 10% of main current through a galvanometer of $99 \Omega ?$ Sol.

$S =\frac{ I _{ g } G }{ I – I _{ g }}$

where $I_{9}=\frac{10}{100} I=0.1 I$

So $S=\frac{0.1 I \times 99}{I(1-0.1)}=\frac{9.9}{0.9}=11 \Omega$.

Moving Coil Galvanometer Class 12
Ever wondered how the utility company detects how much power is used each month? The galvanometer is an instrument used to determine the presence, direction, and strength of an electric current in a conductor. Here we will study about the Galvanometer, moving Coil Galvanometer class 12 and its types. These are instruments used for detection and measurement of small currents.   The moving coil galvanometer work on the principle that when a current carrying coil is placed in a magnetic field it experiences a torque.   The different type of moving coil galvanometers are (a) Pivoted Galvanometer : It consists of a coil of fine insulated wire wound on a metallic frame. The coil is mounted on two jewelled pivots and is symmetrically placed between cylindrical pole pieces of a strong permanent horse-shoe magnet. (b) Dead beat Galvanometer : Here coil is wound over the metallic frame to make it dead beat. On passing current the galvanometer shows a steady deflection without any oscillation. The damping is produced by eddy currents. (c) Ballistic Galvanometer : This is used for measurement of charge. Here coil is wound on an insulating frame and oscillates on passing current. In moving coil galvanometer the deflection produced is proportional to current flowing through galvanometer i.e. they have linear scale of measurement. In equilibrium deflecting torque = restoring torque i.e. NIAB $=\mathrm{C} \theta \quad$ or $\quad I=\left(\frac{C}{N A B}\right) \theta \quad$ or $I=K \theta$ where $\mathrm{K}$ is galvanometer constant.

### Current Sensitivity

This is defined as the deflection produced in galvanometer when a unit current flows through it. Current sensitivity $CS =\frac{\theta}{ I }=\frac{ NAB }{ C }$ radian/ampere or division/ampere (a) The sensitivity can be increased by increasing number of turns in coil (N), area of cross-section of coil (A), magnetic field B and decreasing torsional constant (C). (b) The reciprocal of current sensitivity is called figure of merit. Figure of merit $FM =\frac{ I }{\theta}=\frac{ C }{ NAB }$     Voltage Sensitivity  This is defined as the deflection produced in galvanometer when a unit voltage is applied across its terminals. Voltage sensitivity $VS =\frac{\theta}{ V }=\frac{\theta}{ IR }=\frac{ NAB }{ CR }$ division/volt, where R is resistance of coil.
• Shunting a galvanometer reduces its current sensitivity.
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. Potentiometer 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.
7. ##### measurement of small thermo emf

PRECAUTIONS IN THE USE OF POTENTIOMETER

## Determination of unknown EMF or Potential Difference

1. Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$
2. If unknown emf $\mathrm{E}_{1}$ is balanced at length $\ell_{1}$ then $E_{1}=x \ell_{1}=\left(\frac{E_{0}}{\ell_{0}}\right) \ell_{1}$
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. In state of zero deflection no current flows in galvanometer circuit but it flows in primary circuit.

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

Sensitivity of Potentiometer – Current Electricity || Class 12, JEE & NEET
Sensitivity of Potentiometer means the smallest potential difference it can measure. It can be increased by reducing the potential gradient.
A potentiometer is sensitive if it is capable of measuring the small potential differences and it shows significant change in balancing length for a small change in the potential difference being measured.
Sensitivity of Potentiometer means the smallest potential difference that can be measured by using it and this can be achieved by decreasing the potential gradient by increasing the length of the wire or reducing the current in the potentiometer using rheostat.
1. Smaller the potential difference that can be measured with a potentihttps://www.esaral.com/potentiometer-working-principle-class-12/ometer more is the sensitivity of the potentiometer.
2. The sensitivity of potentiometer is inversely proportional to potential gradient $(S \propto 1 / x)$
3. The sensitivity can be increased by (a) Increasing length of potentiometer wire (b) For a potentiometer wire of fixed length potential gradient is decreased by reducing the current in circuit.

## How Sensitivity of a potentiometer can be increased?

It can be increased by decreasing its potential gradient. This can be done by two ways:

• by increasing the length of potentiometer wire.
• if the potentiometer wire is of fixed length, the potentiometer gradient can be decreased by reducing the current in the potentiometer wire circuit with the help of rheostat and using a single cell.

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

Class 12 Potentiometer Questions for NEET | JEE
Here are the Potentiometer Questions for NEET, IIT JEE and Board Exams.

## Potentiometer Questions for NEET, IIT JEE and Board Exams

Ex. In the primary circuit of a potentiometer a battery of $2 \mathrm{V}$ and a rheostat of $22 \Omega$ are connected. If the length of potentiometer wire is $10 \mathrm{m}$ and its resistance is $18 \Omega$ then find the potential gradient in wire. Sol. The potential gradient $x=\left(\frac{E}{r+R+R_{h}+R_{e}}\right) \frac{R}{L} \quad r=R_{e}=0 \Omega$ So $x=\left(\frac{2}{0+18+22+0}\right) \frac{18}{10}=0.09 V / m$
Ex. No current flows if the terminals of a cell are joined with 125 cm length of potentiometer wire. If a 20 resistance is connected in parallel to cell balancing length is reduced by 25 cm. Find internal resistance of cell. Sol. $r=\frac{\ell_{1}-\ell_{2}}{\ell_{2}} \times R =\frac{125-100}{100} \times 20=5 \Omega$
Ex. Figure shows use of potentiometer for comparision of two resistances. The balance point with standard resistance R = 10 is at 58.3 cm, while that with unknown resistance X is 68.5 cm. Find X.

Sol. Let $E_{1}$ and $E_{2}$be potential drops across R and X so $\frac{E_{2}}{E_{1}}=\frac{1 \times}{1 R}=\frac{x}{R}$ or $X=\frac{E_{2}}{E_{1}} R$ But $\frac{E_{2}}{E_{1}}=\frac{l_{2}}{l_{1}}$ so $X=\frac{\ell_{2}}{\ell_{1}} R=\frac{68.5}{58.3} \times 10=11.75 \Omega$

#### Browse More Topics Related to Potentiometer:

Ex. The length of potentiometer wire is 40cm. Zero deflection in galvanometer is obtained at point F. Find the balancing length AF. Sol. Let x be potential gradient and $V_{1}$ and $V_{2}$ be potential difference across 8$\Omega$ and 12$\Omega$ respectively. $\frac{V_{1}}{V_{2}}=\frac{8}{12}=\frac{x \ell}{x(40-\ell)}$ or $\frac{l}{40-l}=\frac{2}{3}$ or $\ell=16 cm$
Ex. If the current in the primary circuit of a potentiometer wire is 0.2 A, specific resistance of material of wire is $40 \times 10^{-8} \Omega m$ and area of cross-section is $0.8 \times 10^{-6} m ^{2}$ Calculate potential gradient? Sol. Potential gradient $x=\frac{V}{L}=\frac{I R}{L}=\frac{I \rho}{A}=\frac{0.2 \times 40 \times 10^{-8}}{0.8 \times 10^{-6}}=0.1$ volt $/ m$
Ex. A battery of emf 2V is connected in series with a resistance box and a 10m long potentiometer with resistance $1 \Omega / m$. A 10 mV potential difference is balanced across the entire length of wire. Calculate the current flowing in wire and resistance in resistance box. Sol. Potential gradient $x=\frac{E}{r+R+R_{e}}\left(\frac{R}{L}\right)=\frac{2}{0+10+R_{e}} \times 1=\frac{2}{10+R_{e}}$ Potential difference $V=x l$ or $10 \times 10^{-3}=\frac{2}{10+ R _{ e }} \times 10$ or $R_{e}=1990 \Omega$ Current flowing $I=\frac{E}{R+R_{e}}=\frac{2}{10+1990}=1 m A$
Ex. While measuring the potential difference between the terminals of a resistance wire the balance point is obtained at 78.4 cm. The same potential difference is measured as 1.2 V with voltmeter. If standard cell of emf 1.018 V is balanced at 63.2 cm then find error in reading of voltmeter. Sol. $E_{0}=x l_{0}$ and $V=x \ell$ or $V=\frac{E_{0}}{\ell_{0}} \ell$ $V=\frac{1.018 \times 78.4}{63.2}=1.26 volt$ $V=\frac{1.018 \times 78.4}{63.2}=1.26 volt$ so error = 1.2 – 1.26 = –0.06 V

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

Advantages of Potentiometer Over Voltmeter – Current Electricity || Class 12, JEE & NEET
Potentiometer is defined as an instrument which is used for measuring an electromotive force (emf) by balancing emf against the potential difference that is produced by passing known amount of current. Here we will study about the advantages and disadvantages of Potentiometer. Also you’ll get to know about the advantages of Potentiometer over Voltmeter.

1. Potentiometer works on zero deflection method so possibility of error is very small.
2. The standardization of potentiometer can be done directly with a standard cell.
3.  This is highly sensitive so can be used to measure small emf’s.
4.  For more accuracy the length of potentiometer wire can be increased according to requirement.
5.  It does not draw any current from the circuit in which it is used for measurement.

## Advantages of Potentiometer over Voltmeter

1. The use of Potentiometer is inconvenient.
2. The area of cross-Section of potentiometer wire must be uniform which is practically not possible.
3. During experiment the temperature of potentiometer wire must remain uniform. This is difficult because of flow of current.
4. The major disadvantage is that it requires a large force to move their sliding contacts i.e. wiper. There is wear and tear due to movement of the wiper. It reduces the life of this transducer.
5. Also, there is limited bandwidth.

## Conclusion

The Potentiometer and voltmeter both measures the emf in volts. The Potentiometer is used in a circuit where the accurate value of voltage is required. For approximate calculation, the voltmeter is used.

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

Potentiometer Working Principle, Uses, Advantages & Disadvantages || Class 12, JEE & NEET
A potentiometer is a device used to measure the potential difference in a circuit. As we know that potential difference is the amount of work done in bringing a charge from one point to another. When there is potential difference in a circuit, the current flows through the circuit. The Unit of Potential Difference is measured in Volts. The potential difference of a circuit can be measure by a voltmeter. Now we will see Potentiometer Working Principle with the uses of Potentiometer.

## Potentiometer Working Principle

The potential drop across any section of wire of uniform cross-section and composition is proportional to length of that section if a constant current flows through it. If $\mathbf{I}$ is the current in potentiometer wire AB of uniform cross-sectional area A, length L and specific resistance $\rho$ then unknown potential difference across AC is $V =\frac{ I \rho \ell}{ A }$ and known potential difference across AB is $E_{p}=\frac{I \rho L}{A}$
At balance point unknown potential difference = known potential difference or $\frac{V}{\ell}=\frac{E_{p}}{L}$ or $V=\left(\frac{E_{p}}{L}\right) \ell$ or $V=x l$ $$\text { so } \quad V \propto \ell$$ where x = $E_{p} / L$= potential gradient i.e. fall of potential per unit length of potentiometer. Important Points
1. Potentiometer was devised by Poggendorf.
2. The positive terminals of driving cell and unknown cell must be connected to the same end of potentiometer wire to obtain a balance point.
3. A balance point is obtained on potentiometer wire if the fall of potential along potentiometer wire due to driving cell $E_{p}$ is greater than emf of cell to be balanced.
4. The diameter of potentiometer wire must be uniform, specific resistance $P$ must be large and temperature coefficient  of material of wire must be small.
5. It is based of null deflection method i.e. while measuring emf it does not draw any current from source of driving emf.
6. While measuring unknown potential difference the resistance of potentiometer wire becomes infinite.

### Browse More Topics Related to Potentiometer:

The fall of potential per unit length of potentiometer wire is called potential gradient. r = internal resistance of driving cell; $R_{h}$ = resistance of rheostat, $R_{e}$ = external series resistance, R is resistance of potentiometer wire, L is length of potentiometer wire. The current through primary circuit $I=\frac{E_{p}}{r+R_{h}+R_{e}+R}$
#### The potential gradient $x=\frac{I R}{L}=\frac{E_{p}}{r+R_{h}+R_{e}+R}\left(\frac{R}{L}\right)$
1. If $R_{h}$ = 0 and $R_{e}$ = 0 x = $X_{\max }$ = $\frac{E_{p} R}{(r+R) L}0024 2. x =$\mathrm{X}_{\mathrm{min}}$=$\frac{E_{p}}{R+R_{h}+R_{e}+R}\left(\frac{R}{L}\right)$3.$x =\frac{ V }{ L }=\frac{\text { current } \times \text { resistance of potentiometer wire }}{\text { length of potentiometer wire }}= I \left(\frac{ R }{ L }\right)$4. where R/L is resistance per unit length of potentiometer wire. 5.$R=\frac{\rho L}{A}$or$\frac{ R }{ L }=\frac{\rho}{ A }$so$x =\frac{ I \rho}{ A }=\frac{ I \times \text { specific resistance of material }}{\text { area of cross – sec tion }}$6. Unit of potential gradient is volt/meter and dimensions are$M^{1} L^{1} T^{-3} A^{-1}$. 7. The potential gradient depends only on primary circuit and is independent of secondary circuit. 8. On increasing the temperature of potentiometer wire there is no change in potential gradient if a constant current is maintained. If current is altered due to change in resistance of wire then potential gradient will change. 9. Keeping the thickness of potentiometer wire constant if the length is changed from$L_{1}$to$L_{2}$then ratio of potential gradient will be$\frac{x_{1}}{x_{2}}=\frac{L_{2}}{L_{1}}$10. If two wires of length$\mathrm{L}_{1}$and$\mathrm{L}_{2},$resistances$\mathrm{R}_{1}$and$\mathrm{R}_{2}$are joined in series with a battery of emf$E_{p}$and a rheostat than the ratio of potential gradients can be calculated as$x_{1}=\left(\frac{E_{p}}{R_{1}+R_{2}}\right) \frac{R_{1}}{L_{1}} \quad$and$\quad x_{2}=\left(\frac{E_{p}}{R_{1}+R_{2}}\right) \frac{R_{2}}{L_{2}} \quad$or$\quad \frac{x_{1}}{x_{2}}=\frac{R_{1}}{R_{2}} \cdot \frac{L_{2}}{L_{1}}$The potential gradient depends on: 1. emf of battery in primary circuit$\left(E_{p}\right)$and its internal resistance (r). 2. Length of Potentiometer wire (L), its radius and its resistance (R). 3. Specific resistance of material of wire$(\rho)$. 4. Current flowing through the wire. 5. Additional resistance like resistance of rheostat$\left(R_{h}\right)$and series external resistance$\left(\mathrm{R}_{\mathrm{e}}\right)\$.