CBSE Class 12 Physics Electrical Instrument Theory and Examples

CBSE Class 12 Physics Electrical Instrument Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

ELECTRICAL INSTRUMENTS

ELECTRICAL INSTRUMENTS

1. MOVING-COIL METERS

1.1 Galvanometer

The instrument used to measure the strength of a current, by measuring the deflection of a coil due to the torque produced by a magnetic field, is known as a galvanometer.

Types of galvanometer

1. Moving magnet type: in this type of galvanometer the magnet is movable and the coil is stationary.
2. Moving coil type: in this type of galvanometer the coil is movable and the magnet is stationary.

(a) Suspended coil type galvanometer

1. The coil is suspended by a fine phosphor bronze fibre.
2. The deflection of the coil is measured by a lamp and scale arrangement.
3. Its sensitivity is high (a current of the order of \(10^{-9}\) A can be measured by it).

(b) Pivoted coil type galvanometer

1. The coil is pivoted on a pivot about which it is free to rotate.
2. The deflection of the coil is determined by a pointer rotating on a circular scale.
3. Its sensitivity is low (a current of the order of \(10^{-6}\) A can be measured).

1.2 Moving coil type galvanometer

1. Principle: when a current-carrying coil is placed in a magnetic field, it experiences a couple and the coil rotates.

(b) The torque of this couple is given by

\[\tau=NiAB\sin\alpha\]

where \(N\) is the number of turns in the coil, \(i\) is the current through the coil, \(B\) is the intensity of the magnetic field and \(\alpha\) is the angle between the magnetic field and the plane of the coil.

(c) For a radial magnetic field \(\alpha=90^\circ\), therefore

\[\tau=NiAB\]

(d) In the state of equilibrium of the coil, deflecting torque = restoring torque, so

\[NiAB=C\theta\]

where \(C\) is the restoring torque per unit deflection (it depends on the elastic properties of the spring). Hence

\[i=\frac{C}{NAB}\,\theta\quad\text{or}\quad i\propto\theta\]

The current is proportional to the deflection. The constant \(\dfrac{C}{NAB}\) is called the galvanometer constant.

(e) Sensitivity of a galvanometer

1. The deflection of the galvanometer coil per unit current is defined as the sensitivity (current sensitivity) of the galvanometer.
2. \(S_i=\dfrac{\theta}{i}\)
3. The deflection obtained by unit current is \(S_i\).

(f) The galvanometer which gives more deflection for a small current is more sensitive. To increase the sensitivity, the values of \(N\), \(A\), \(B\) must be large and the value of \(C\) must be small.

(g) Figure of merit (K)

1. The reciprocal of the current sensitivity is defined as the figure of merit of the galvanometer.
2. The current required for unit deflection in the galvanometer is known as the figure of merit \((K)\).
3. \(S_i=\dfrac1K\) or \(K=\dfrac1{S_i}\)

(h) Voltage sensitivity (S_v)

1. The deflection of the galvanometer coil per unit voltage is defined as the voltage sensitivity of the galvanometer.
2. \(S_v=\dfrac{\theta}{V}=\dfrac{\theta}{iR_G}=\dfrac{NAB}{CR_G}\), where \(R_G\) is the resistance of the galvanometer coil.

Examples: Moving-coil meters

Which is more sensitive, a suspended type galvanometer or a pivoted type galvanometer?

Answer: Suspended type galvanometer.

The sensitivity of a suspended type galvanometer is of the order of \(10^{-9}\) A, while that of a pivoted type is of the order of \(10^{-6}\) A.

In your practical classes, which type of galvanometer do you use?

Answer: Pivoted-coil (Weston type) galvanometer.

In the suspended coil galvanometer the suspension strip is very delicate and may be broken by a small jerk. Therefore this galvanometer cannot be taken from one place to another. This difficulty has been removed by the Weston galvanometer.

Two moving coil meters \(m_1\) and \(m_2\) have the following particulars: \(R_1=10\ \Omega\), \(N_1=30\), \(A_1=3.6\times10^{-3}\ \text{m}^2\), \(B_1=0.25\ \text{T}\); \(R_2=14\ \Omega\), \(N_2=42\), \(A_2=1.8\times10^{-3}\ \text{m}^2\), \(B_2=0.50\ \text{T}\). (The spring constants are identical for the two meters.) The ratios of the current sensitivity and voltage sensitivity of \(m_2\) and \(m_1\) are respectively:

1. 1.4, 1
2. 0.93, 1
3. 1.4, 1.4
4. 1.4, 0.93

Answer: (A)

\[\frac{\text{Current sensitivity of }m_2}{\text{Current sensitivity of }m_1}=\frac{N_2A_2B_2}{N_1A_1B_1}=\frac{42\times1.8\times10^{-3}\times0.5}{30\times3.6\times10^{-3}\times0.25}=1.4\]

\[\frac{\text{Voltage sensitivity of }m_2}{\text{Voltage sensitivity of }m_1}=\frac{\text{Current sensitivity of }m_2}{\text{Current sensitivity of }m_1}\times\frac{R_1}{R_2}=1.4\times\frac{10}{14}=1\]

1.3 Difference between a ballistic galvanometer and a dead-beat galvanometer

Shunt

The small resistance connected in parallel to a galvanometer coil, in order to control the current flowing through the galvanometer, is known as a shunt.

Merits of a shunt

1. To protect the galvanometer coil from burning.
2. It can be used to convert any galvanometer into an ammeter of the desired range.

Demerits of a shunt

A shunt resistance decreases the sensitivity of the galvanometer.

2. AMMETER

1. An ammeter is a shunted galvanometer which is used to measure current in a circuit.
2. It is always connected in series so that the entire current passes through it.
3. In principle, the current in the circuit must not change when a current-measuring device like an ammeter is introduced in the circuit; therefore an ideal ammeter must have zero resistance.
4. However, in practice a moving coil meter has some resistance. Due to this the current in the circuit is modified (reduced) when a moving coil ammeter is connected in the circuit.

3. VOLTMETER

A voltmeter is a device used to measure the potential difference across two points in an electrical circuit. It is connected in parallel to these points.

Note: the ammeter is connected in series to a circuit, while the voltmeter is connected in parallel to the circuit.

Conversion of a galvanometer into an ammeter

(a) A galvanometer is converted into an ammeter of a given range by connecting a low shunt \(S\) in parallel to it.

(b) From the figure

\[I_gG=(I-I_g)S\quad\text{or}\quad S=\frac{I_gG}{I-I_g}\]

(c) The higher the range \(I\) of the ammeter, the lower is the value of \(S\) required for the conversion of the galvanometer into an ammeter.

(d) Let a galvanometer of resistance \(G\) be connected in series with a resistance \(R\) and a cell of e.m.f. \(E\); let \(n\) be the deflection (in divisions) and \(N\) the total number of divisions on the scale of the galvanometer. Then

\[I_g=\frac{E}{R+G}\left(\frac Nn\right)\]

The effective resistance of the ammeter is

\[R_A=\frac{GS}{G+S}\qquad\text{For an IDEAL AMMETER, }R_A=0.\]

Conversion of a galvanometer into a voltmeter

(a) A galvanometer is converted into a voltmeter by connecting a high resistance \(R\) in series with the galvanometer.

(b) From the figure

\[I_g(R+G)=V\quad\text{or}\quad R=\frac V{I_g}-G\]

(c) The higher the range \(V\) of the voltmeter, the higher is the value of \(R\) required for the conversion of the galvanometer into a voltmeter.

(d) The effective resistance of the voltmeter is

\[R_V=R+G\qquad\text{For an IDEAL VOLTMETER, }R_V=\infty.\]

Examples: Ammeter and voltmeter

A voltmeter, an ammeter and a resistance are connected in series to a lead accumulator. There is a deflection in the voltmeter but the deflection in the ammeter is negligible. Why?

Answer: The voltmeter being in series, the resistance of the circuit becomes very high and so the current is very low.

This current, on passing through the coil of the voltmeter, produces some deflection, but in the ammeter most of it goes through the shunt of the ammeter and so the current going through the coil of the ammeter is too small to produce any deflection.

A galvanometer has a coil of resistance 60 Ω and shows a full-scale deflection for 50 μA current. To convert this galvanometer into an ammeter of range 10 mA, the required shunt resistance will be:

1. 0.30 Ω
2. 0.20 Ω
3. 0.6 Ω
4. 0.40 Ω

Answer: (A)

Given \(I_g=50\ \mu\text{A}=50\times10^{-6}\) A, \(I=10\ \text{mA}=10\times10^{-3}\) A, \(G=60\ \Omega\).

\[S=\frac{I_gG}{I-I_g}=\frac{50\times10^{-6}\times60}{10\times10^{-3}-50\times10^{-6}}=0.30\ \Omega\]

A galvanometer has a coil of resistance 100 Ω and shows a full-scale deflection when 100 μA current is passed through it. The resistance required to use it as a voltmeter of range 5 V will be:

1. 95 kΩ
2. 9.5 kΩ
3. 49.9 kΩ
4. 4.99 kΩ

Answer: (C)

Given \(I_g=100\ \mu\text{A}=100\times10^{-6}\) A, \(G=100\ \Omega\) and \(V=5\) V.

\[R=\frac V{I_g}-G=5\times10^{4}-100=49.9\ \text{k}\Omega\]

In the figures, \(R_1=R_2=200\ \Omega\) and the e.m.f. of the source is 40 V. The resistance of voltmeter \(V_1\) is \(R_{V_1}=2\ \text{k}\Omega\) and that of voltmeter \(V_2\) is \(R_{V_2}=20\ \text{k}\Omega\). What is (i) the p.d. across A, B in fig. (a)? (ii) the readings of voltmeters \(V_1\) and \(V_2\) in figs. (b) and (c)?

Solution. (i) In fig. (a) the current in the circuit is \(i=\dfrac{40}{400}=0.1\) A, so the p.d. between A and B is \(V_{AB}=0.1\times200=20\) V.

(ii) In fig. (b) the effective resistance between A and B is

\[R_1'=\frac{200\times2000}{2200}=\frac{2000}{11}\ \Omega\]

The current in the circuit is \(i'=\dfrac{40}{4200/11}\) A, so the p.d. across A, B is

\[V'_{AB}=\frac{2000}{11}\times\frac{40\times11}{4200}=19.05\ \text{V}\]

Note: the voltmeter does not read 20 V but reads only 19.05 V. The error in the reading = voltmeter reading − actual p.d. = −0.95 V.

In fig. (c) the effective resistance between A and B is \(R_1''=\dfrac{200\times20000}{20200}\ \Omega\). Proceeding in the same manner, the p.d. between A and B is

\[V''_{AB}=I''\times R_1''=\frac{40\times101}{40200}\times\frac{20000}{101}=19.90\ \text{V}\]

Note: again the voltmeter does not read 20 V, but the error is much reduced (error = −0.1 V). Thus the higher the resistance of the voltmeter, the lower is the error in the measurement of the p.d.

The above example illustrates that there always exists an error in the reading of a moving coil voltmeter. In principle the error would reduce to zero when \(R_V=\infty\). But such a voltmeter will not give any deflection and thus cannot be used for measurements. One device which behaves as an ideal voltmeter is the potentiometer.

A galvanometer having 30 divisions has a current sensitivity of 20 μA/division. It has a resistance of 25 Ω. To convert it into an ammeter measuring up to 1 ampere and a voltmeter reading up to 1 volt, the resistances required will be respectively: [ROORKEE-87]

1. 0.015 Ω, 0.985 Ω
2. 0.012 Ω, 0.985 Ω
3. 0.985 Ω, 0.015 Ω
4. 0.985 Ω, 0.012 Ω

Answer: (A)

The current \(i_g\) for a full-scale deflection in the galvanometer is \(i_g=20\ \mu\text{A}\times30=6\times10^{-4}\) A.

Let \(S\) be the shunt resistance (in parallel) to convert it into an ammeter of range \(i\) ampere. Then \(i_g=\dfrac{S}{S+G}\,i\), where \(G\) is the resistance of the galvanometer. Putting in the values,

\[6\times10^{-4}=\frac{S}{S+25}\times1\;\Rightarrow\;S=0.015\ \Omega\]

Suppose the resistance of the converted ammeter is \(r\); then \(\dfrac1r=\dfrac1{25}+\dfrac1{0.015}\Rightarrow r\approx0.015\ \Omega\).

To convert the 1 A ammeter into a voltmeter of 1 V, the necessary series resistance is

\[R=\frac Vi-r=\frac11-0.015=0.985\ \Omega\]

4. POTENTIOMETER

1. It is an ideal instrument for measuring the e.m.f. of a cell or the potential difference between two points of an electric circuit.
2. It is equivalent to a voltmeter with infinite resistance because, when it is used in an electric circuit, it does not draw any current from the main circuit. That is why it is called an 'ideal voltmeter'.
3. A potentiometer consists of a long wire (1 m to 10 m long) of uniform cross-section and homogeneous material.
4. The potentiometer wire is made of a material for which the temperature coefficient of resistance is low and the resistivity is high.
5. Materials used to construct a potentiometer are alloys, e.g. eureka, manganin, constantan, etc.

(f) Principle: the potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

[\(E_P\) (or \(E_B\)) is the e.m.f. of the cell used in the primary circuit; \(E_S\) (or \(E\)) is the e.m.f. of the cell used in the secondary circuit; \(R_W\) (or \(R_P\)) is the resistance of the potentiometer wire.]

1. In the balanced state it does not draw any current from the main circuit whose p.d. is being measured.
2. If the galvanometer does not show any deflection, this is possible only if the p.d. between A and B = the p.d. between X and Y.
3. Similarly, when we connect the unknown p.d. between X and Y and the galvanometer does not show any deflection, then the p.d. across A and C = the p.d. across X and Y.
4. Also at this stage no current flows from X to A or A to X or C to Y or Y to C, i.e. the primary circuit and the secondary circuit work independently.

(g) Potential gradient (\(\phi\))

1. The fall of potential per unit length of the potentiometer wire is known as the potential gradient. From the above figure \(\phi=\dfrac{E_P}{L}\).

(ii) \(\phi\) directly depends upon:

1. the resistance per unit length of the potentiometer wire \(\rho\) \(\left(\therefore\ \rho=\dfrac{R_W}{L}\right)\)
2. the radius of the potentiometer wire (\(a\))
3. the specific resistance of the material of the potentiometer wire (\(R\))
4. the current flowing through the potentiometer wire.

(iii) \(\phi\) indirectly depends on:

1. the e.m.f. of the battery in the primary circuit (\(E_P\))
2. the resistance of the rheostat in the primary circuit (\(R_h\))
3. the total resistance of the potentiometer wire \(R\) and its total length (\(\therefore\ R_W=\rho L\))

(iv) Various formulae for the potential gradient:

(A) When no resistance other than the potentiometer wire is connected in the potentiometer circuit:

\[\phi=\frac{E_P}{L}\]

\(E_P\) = e.m.f. of the battery used in the primary circuit; \(L\) = length of the potentiometer wire.

(B) When an additional resistance is connected in the circuit:

\[\phi=\frac{E_PR_W}{L\,(R_W+R_h+r)}\]

\(R_W\) = resistance of the potentiometer wire; \(R_h\) = resistance of the rheostat; \(r\) = internal resistance of the battery.

(C) When the current flowing in the primary circuit is given:

\[\phi=I\rho\quad\text{where }\rho=\frac{R_W}{L}\]

\(\rho\) = resistance per unit length of the potentiometer wire.

(D) When the specific resistance \(k\) of the potentiometer wire is given:

\[\phi=\frac{Ik}{A}=\frac{Ik}{\pi a^2}\]

where \(A\) = area of cross-section and \(a\) = radius of the potentiometer wire.

(E) When the potential difference \(V\) is constant:

\[\frac{\phi_1}{\phi_2}=\frac{L_2}{L_1}\]

(F) When two wires of lengths \(L_1\) and \(L_2\) and resistances \(R_1\) and \(R_2\) are joined together to form the potentiometer wire, then

\[\frac{\phi_1}{\phi_2}=\frac{R_1}{R_2}\cdot\frac{L_2}{L_1}\]

(G) Maximum potential gradient:

\[\phi_{max}=\frac{E_PR_W}{L\,(R_W+r)}\qquad\text{If }r\to0,\ \phi_{max}=\frac{E_P}{L}\]

(H) Minimum potential gradient:

\[\phi_{min}=\frac{E_PR_W}{L\,(R_W+R_h+r)}\]

(I)

\[\frac{\phi_{max}}{\phi_{min}}=\frac{R_W+R_h+r}{R_W+r}\]

(V) (A) If the length of the potentiometer wire and the potential difference across its ends are kept constant and its diameter is changed from \(d_1\) to \(d_2\), then the potential gradient remains unchanged, i.e. \(\phi_1=\phi_2\).

(B) If the specific resistance of the material of the wire is changed from \(R_1\) to \(R_2\), then also the potential gradient remains unchanged.

(C) The unit of potential gradient is volt/metre or volt/cm.

(D) The dimensions of \(\phi\) are \(MLT^{-3}A^{-1}\).

(h) Standardization of the potentiometer: the process of determining the potential gradient experimentally is known as standardization of the potentiometer. A standard cell \(E_s\) is connected in the secondary circuit and the balancing length \(l\) found; then \(\phi=E_s/l\).

(i) Sensitivity

1. A potentiometer is said to be more sensitive if it measures a small potential difference more accurately.
2. The sensitivity of a potentiometer is assessed by its potential gradient. The sensitivity is inversely proportional to the potential gradient.
3. In order to increase the sensitivity of the potentiometer: (A) the resistance in the primary circuit will have to be increased; (B) the length of the potentiometer wire will have to be increased so that the length may be measured more accurately.

(j) Difference between a potentiometer and a voltmeter

Examples: Potentiometer

In an experiment for the calibration of a voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of the potentiometer wire. The potential difference across the ends of a resistance is found to balance against 220 cm of the wire. The corresponding reading of the voltmeter is 0.5 volt. The error in the reading of the voltmeter will be:

1. −0.15 volt
2. 0.15 volt
3. 0.5 volt
4. −0.05 volt

Answer: (D)

\[\phi=\frac{E_P}{L}=\frac{1.1}{440}=0.0025\ \text{V/cm}\]

Potential difference across \(R=0.0025\times220=0.550\) volt.

Error in the reading of the voltmeter = reading of voltmeter − reading of potentiometer \(=0.5-0.55=-0.05\) volt.

A cell of e.m.f. 2 V and internal resistance 1.5 Ω is connected to the two ends of a wire of length 100 cm. The resistance of the wire is 0.005 Ω/cm. The potential gradient of the wire is:

1. 0.005 V/m
2. 0.5 V/m
3. 0.05 V/m
4. 0.005 V/cm

Answer: (D)

\[\phi=\frac{E_P}{L}\cdot\frac{R_W}{R_W+r}\]

It is given that \(E_P=2\) V, \(L=100\) cm, \(r=1.5\ \Omega\), \(R_W=0.005\times100=0.5\ \Omega\). Thus \(\phi=\dfrac{2\times0.5}{100\,(1.5+0.5)}=0.005\) V/cm.

(Note: \(0.005\) V/cm is the same as \(0.5\) V/m, so option (B) expresses the same value; the key follows option (D).)

If the current in the primary circuit of a potentiometer wire is 0.2 A, the specific resistance of the material of the wire is \(40\times10^{-8}\ \Omega\,\text{m}\) and the area of cross-section of the wire is \(8\times10^{-7}\ \text{m}^2\), the potential gradient will be:

1. 0.1 V/m
2. 0.001 V/m
3. 0.1 V/cm
4. 0.01 V/m

Answer: (A)

\[\phi=I\left(\frac{R_W}{L}\right)=I\rho\frac{1}{A}\quad\left[\because\ R_W=\rho\frac LA\right]\]

\[\phi=\frac{0.2\times40\times10^{-8}}{8\times10^{-7}}=0.1\ \text{V/m}\]

5. APPLICATIONS OF THE POTENTIOMETER

Complete list of applications, in the PDF's numbering: 5.1 Comparison of e.m.f.'s; 5.2 Measurement of internal resistance; 5.3 Measurement of current; 5.4 Comparison of resistances; 5.5 Measurement of small emf – thermo emf. 5.6 Calibration of a voltmeter; 5.7 Calibration of an ammeter. (The shorter list below omits the thermo emf item and so is numbered 5.5 and 5.6 for the calibration entries; the numbering here follows the PDF.)

• 5.1 Comparison of e.m.f.'s
• 5.2 Measurement of internal resistance
• 5.3 Measurement of current
• 5.4 Comparison of resistances
• 5.5 Calibration of a voltmeter
• 5.6 Calibration of an ammeter

5.1 Comparison of e.m.f.'s

Let \(l_1\) and \(l_2\) be the balancing lengths with the cells \(E_1\) and \(E_2\) respectively; then \(E_1\propto l_1\) and \(E_2\propto l_2\), so

\[\frac{E_1}{E_2}=\frac{l_1}{l_2}\]

Note: let \(E_1>E_2\) and both be connected in series.

(a) For this combination, let \(l_1\) be the balancing length; then \(E_1+E_2=\phi\,l_1\) (fig. (a)).

(b) For this combination, let \(l_2\) be the balancing length; then \(E_1-E_2=\phi\,l_2\) (fig. (b)).

\[\text{(c)}\quad\frac{E_1}{E_2}=\frac{l_1+l_2}{l_1-l_2}\]

5.2 Measurement of internal resistance

(a) The balance point (\(l_1\)) is found with the key \(K_2\) open; this measures \(E\), since no current is flowing from the cell: \(E=\phi\,l_1\).

(b) The key \(K_2\) is now closed and a second balance point (\(l_2\)) is found. This measures the p.d. (\(V\)) between the terminals of the cell (in this case some current flows through the resistor, so \(V<E\)): \(V=\phi\,l_2\).

(c) Since \(E=I(r+R)\),

\[V=IR=\frac{ER}{r+R}\;\Rightarrow\;r=\frac{E-V}{I}\;\text{or}\;r=\frac{E-V}{V}\,R=\left(\frac{l_1-l_2}{l_2}\right)R\]

Note: the internal resistance of a cell depends on:

1. the concentration of the electrolyte (the internal resistance decreases as the concentration increases, because more ions are available to carry the current);
2. the distance \(d\) between the electrodes (\(r\propto d\));
3. the area of the electrodes \(A\) dipped in the electrolyte (\(r\propto\dfrac1A\)).

5.3 Measurement of current

(a) First the balancing length (\(l_1\)) is determined for the standard cell \(E_S\): \(E_S=\phi\,l_1\) ...(i). This determines the potential gradient \(\phi\).

(b) Next the potential drop \(V=IR\) across a known resistance is balanced (balancing length \(l_2\)). Then \(V=IR=\phi\,l_2\) ...(ii).

From (i) and (ii):

\[I=\frac{l_2}{l_1}\cdot\frac{E_S}{R}\]

Usually \(R=1\ \Omega\) is taken; then \(I=\dfrac{l_2}{l_1}E_S\).

5.4 Comparison of resistances

(a) If the balancing lengths for \(R_1\) and \(R_2\) are respectively \(l_1\) and \(l_2\), then

\[\frac{R_1}{R_2}=\frac{l_1}{l_2}\]

(b) Let the balancing length for the resistance \(R_1\) (when x and y are connected) be \(l_1\), and for \(R_1+R_2\) (when y and z are connected) be \(l_2\). Then \(IR_1=\phi\,l_1\) and \(I(R_1+R_2)=\phi\,l_2\), so

\[\frac{R_2}{R_1}=\frac{l_2-l_1}{l_1}\]

5.5 Calibration of a voltmeter

(a) Calibration means checking the correctness of the readings obtained from a voltmeter. [Errors in these may arise due to non-uniformity of the markings on the scale, non-uniformity of the spring constant or zero error.]

(b) To calibrate a voltmeter, the potential difference between points C and D is measured both by the voltmeter (\(V\)) and by the potentiometer (\(V'\)). Let \(l_1\) and \(l_2\) be the balancing lengths for the standard cell and for the p.d. across CD respectively; then

\[V'=\frac{l_2}{l_1}\,E_S\]

(c) The error in the voltmeter reading = voltmeter reading (\(V\)) − potentiometer reading (\(V'\)), i.e. \(e=V-V'\).

(d) A graph between the error (\(e\)) and the voltmeter reading (\(V\)) is called a calibration curve. It is obtained by varying the potential difference between C and D by adjusting the resistance in a resistance box (R.B.) and repeating the measurement of \(V\) and \(V'\) for different scale markings of the voltmeter. The curve is usually zig-zag.

5.6 Calibration of an ammeter

(A) The current in the secondary circuit is measured by the ammeter (A) and also by the potentiometer. The potential difference across a 1 Ω coil is \(V'=I'\times1=\phi\,l_2\) ...(i), where \(l_2\) is the balancing length for the p.d. across points C and D.

(B) If \(l_1\) is the balancing length for the standard cell \(E_S\), then \(E_S=\phi\,l_1\) ...(ii).

(C) From (i) and (ii): \(I'=\dfrac{l_2}{l_1}E_S\).

(D) The ammeter reading (\(I\)) and the potentiometer reading (\(I'\)) are used to determine the error: \(e=I-I'\).

(E) The curve between \(e\) and \(I\) usually comes out zig-zag.

SOME COMMENTS

(a) Consider two wires AB and BC joined at B. The radii are \(r_1\) and \(r_2\) and the lengths are \(L_1\) and \(L_2\). They are connected by a cell of e.m.f. \(E\) as shown; the ratio of the potential gradient on the AB portion and the BC portion is

\[\frac{\phi_{AB}}{\phi_{BC}}=\frac{I(R_{AB}/L_1)}{I(R_{BC}/L_2)}=\frac{R_{AB}}{R_{BC}}\times\frac{L_2}{L_1}=\frac{\rho_1}{\rho_2}\times\frac{r_2^2}{r_1^2}\]

where \(\rho_1\) and \(\rho_2\) are the specific resistances of the materials of the two wires.

(b) Let \(E_P\) be the e.m.f. of the primary battery and \(L\) the length of the potentiometer wire; the balancing length for an unknown e.m.f. \(E\) is \(l\). Keeping \(E_P\) the same, if the length of the potentiometer wire is changed to \(L'\), the balancing length for the unknown e.m.f. \(E\) also changes. The new balancing length is

\[l'=\frac E{\phi'}=\frac{E}{E_P/L'}\quad\text{But }E=\phi\,l=\frac{E_P}{L}\,l\;\Rightarrow\;l'=\left(\frac{L'}{L}\right)l\]

(c) Consider the circuit shown in the figure (three resistors \(R_1\), \(R_2\), \(R_3\) in series with a cell, and a cell \(E_1\) with key K connected across \(R_2\)). If on switching on the key K the current in the circuit remains the same, what is the e.m.f. \(E_1\)? Obviously, the p.d. across the points C and D is equal to the e.m.f. \(E_1\): \(E_1=IR_2\), where the current \(I\) in the circuit is \(I=\dfrac E{R_1+R_2+R_3}\). Thus

\[E_1=\frac{ER_2}{R_1+R_2+R_3}\]

In the following circuit a 10 m long potentiometer wire of resistance 1.2 ohm/m, a resistance \(R_1\) and an accumulator of e.m.f. 2 V are connected in series. When the e.m.f. of a thermocouple is 2.4 mV, the deflection in the galvanometer is zero (the balancing length is 5 m). The current supplied by the accumulator will be:

1. \(4\times10^{-4}\) A
2. \(8\times10^{-4}\) A
3. \(4\times10^{-3}\) A
4. \(8\times10^{-3}\) A

Answer: (A)

\(E=\phi\,l=I\rho\,l\), therefore

\[I=\frac{E}{\rho\,l}=\frac{2.4\times10^{-3}}{1.2\times5}=4\times10^{-4}\ \text{A}\]

The graph between the error in the voltmeter reading and the actual voltmeter reading is known as:

1. calibration curve
2. exponential curve
3. logarithmic curve
4. cycloid

Answer: (A) Calibration curve

The graph between the error in the voltmeter reading and the actual voltmeter reading is usually:

1. parabolic
2. elliptical
3. zig-zag
4. straight line

Answer: (C) It is a zig-zag curve.

It is obtained by varying the potential difference between points C and D by adjusting the resistance in the resistance box (R.B.) and repeating the measurement of \(V\) and \(V'\) for different scale markings of the voltmeter.

In an experiment for the calibration of a voltmeter, a standard cell of e.m.f. 1.1 V is balanced across 5.50 m length of the potentiometer. If a current of 0.1 A passes through a 2.3 Ω resistance, then the balancing length for the p.d. across this resistance will be:

1. 1.15 m
2. 11.5 m
3. 11.5 cm
4. 0.23 m

Answer: (A)

From the first data, \(\phi=\dfrac{1.1}{5.5}=\dfrac15\) V/m.

From the second data, \(V=0.1\times2.3=0.23\) V.

\[l=\frac V\phi=\frac{0.23}{1/5}=1.15\ \text{m}\]

In the circuit shown (resistors of 6 Ω, 8 Ω and 10 Ω in series with a 12 V cell), the current distribution remains unchanged on connecting the cell \(E\) across the 8 Ω resistor. What is the value of \(E\)?

1. 12 volt
2. 6 volt
3. 4 volt
4. 2 volt

Answer: (C)

\[I=\frac{E_P}{R_1+R_2+R_3}=\frac{12}{6+8+10}=0.5\ \text{A}\]

\[V_2=IR_2=0.5\times8=4\ \text{V}\]

For the current distribution to remain unchanged, \(V_2=E\); hence \(E=4\) V.

The resistance of a potentiometer wire of length 10 m is 20 ohm. A resistance box and a 2 volt accumulator are connected in series with it. The resistance required to be introduced in the box to have a potential drop of one microvolt per millimetre of the potentiometer wire will be:

1. 3880 Ω
2. 3980 Ω
3. 2500 Ω
4. 3600 Ω

Answer: (B)

Here the resistance of the potentiometer wire \(R=20\ \Omega\). Required potential drop along the potentiometer wire \(=1\ \mu\text{V mm}^{-1}\). Length of the potentiometer wire \(l=10\ \text{m}=10^4\) mm.

Therefore the potential drop along the potentiometer wire \(V=1\times10^4\ \mu\text{V}=10^4\ \mu\text{V}=10^{-2}\) V. If \(I\) is the current through the potentiometer wire, then

\[I=\frac VR=\frac{10^{-2}}{20}=5\times10^{-4}\ \text{A}\]

If \(R'\) is the resistance that should be introduced in the resistance box, then

\[I=\frac{E}{R+R'}\;\Rightarrow\;5\times10^{-4}=\frac{2}{20+R'}\;\Rightarrow\;R'=3980\ \Omega\]

6. ADVANTAGES AND DISADVANTAGES OF THE POTENTIOMETER

POINTS TO REMEMBER

1. The principle of the potentiometer is to find out a potential difference by comparing it with a known potential difference.
2. The wire of a potentiometer is made of manganin.
3. The thermal coefficient of resistance of a potentiometer wire is negligible.
4. The value of the specific resistance of the metal of a potentiometer wire is high.
5. The potentiometer is standardized by a cadmium cell.
6. The potentiometer is mainly used for the measurement of potential difference.
7. The potentiometer is an ideal apparatus because it measures without drawing any current.
8. On increasing the temperature of the potentiometer wire, the value of its specific resistance increases.
9. The effective resistance of a potentiometer is infinite.
10. The rate of variation of deflection depends upon the magnitude of the deflection itself, and so does the accuracy. The tangent galvanometer is most accurate when recording a deflection of 45°, and the accuracy decreases for higher and lower values of deflection.
11. A suspended coil galvanometer can measure currents of the order of \(10^{-9}\) ampere.
12. \(I_g\) is the current for full-scale deflection. If the current for a deflection of one division on the galvanometer scale is \(k\) and \(N\) is the total number of divisions on one side of the zero of the galvanometer scale, then \(I_g=k\times N\).
13. A ballistic galvanometer is a specially designed moving coil galvanometer, used to measure the charge flowing through the circuit in small time intervals.
14. In an oscillatory moving coil galvanometer the coil should be wound over a non-metallic frame.
15. In a dead-beat galvanometer the coil is wound on a metallic frame.

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