## Conversion of Galvanometer into Voltmeter Experiment (Class 12)

CONVERSION OF GALVANOMETER INTO VOLTMETER EXPERIMENT (CLASS 12)

Aim: To convert a pointer type galvanometer of known resistance and figure of merit into a voltmeter to read 1 mV per scale division and to calibrate it.

Apparatus: An accumulator, pointer type galvanometer of known resistance and figure of merit, a resistance box of range 1 to 10,000Ω, a high resistance rheostat, a standard millivoltmeter, key etc.

Theory: A galvanometer can be converted into a voltmeter to read any desired value of voltage V per scale division, a suitable high resistance R is connected in series with it. If ig is the current sensitiveness (Figure of merit) and G is resistance of the galvanometer, then,
V = ig(R +G)
Therefore, R = (V — igG)/ ig

Procedure:

(i) To convert the galvanometer into voltmeter

The galvanometer resistance G and its figure of merit ig are noted. The high resistance R to be connected in series with the galvanometer to convert it into a voltmeter to read 1 mV/division is calculated by the equation,
R = (V — igG)/ig, where V = 1 mV = 10-3 V.
A resistance RΩ is unplugged in the resistance box and it is connected in series with the galvanometer. Now the galvanometer is converted into a voltmeter to read 1 mV/div.

(ii) To calibrate the 'galvanometer converted' voltmeter

Connections are made as shown in the figure. The rheostat is used as a voltage divider. V is a standard millivoltmeter.
The rheostat is adjusted to get a defection of 2 divisions in the galvanometer-converted voltmeter. Now its reading is V0 = 2 mV. The standard voltmeter reading V is noted. The correction in the converted voltmeter reading, V — Vo, is calculated.
The experiment is repeated for different values of the converted voltmeter readings. A calibration curve is drawn with the correction V — Vo, along the y-axis and Vo, along the x-axis.

(i) To convert the galvanometer into voltmeter

G = ..... Ω

ig = ..... A/div

V = 1 mV/division

R = (V - igG)/ig = ..... Ω

(ii) To calibrate the converted voltmeter:

 Deflection in galvanometer Converted voltmeter Vo (mA) Standard voltmeter V (mA) Correction, V – Vo(mA) 2 2 3 3 4 4 5 5 - - - -

Result

(i) The galvanometer is converted into a voltmeter to read 1 mV per scale division
(ii) The calibration curve is drawn.

Note:

(i) The correction (V — V0) may be positive or negative
(ii) The calibration curve is drawn by joining the points on the graph with straight lines.

Viva Questions for Conversion of Galvanometer into Ammeter and Voltmeter:

I. What is a Voltmeter?

Voltmeter is a device used to measure p.d. between two points

2. How is a galvanometer converted into a voltmeter?

The galvanometer is converted into a voltmeter by connecting a high resistance in series with the galvanometer.

3. Define resistance of an ideal voltmeter?

The resistance of an ideal voltmeter is infinity.

4. Distinguish between voltmeter and voltameter.

Voltmeter is a device used to measure pal. between two points whereas voltameter is used to carry the process of electrolysis.

5. Define e.m.f. of a cell.

The e.m.f. of a cell is the maximum p.d. between two electrodes of a cell once the cell is in open circuit.

6. Define terminal potential difference.

The potential difference between the two electrons of a cell in a closed circuit is called terminal potential difference. The terminal potential difference should always be less than the electromotive force of the cell.

7. Define relaxation time.

Relaxation time is the time interval between two successive collisions of a ‘free’ electron with the atoms in a metal (10-14s).

## Conversion of Galvanometer into Ammeter Experiment (Class 12)

CONVERSION OF GALVANOMETER INTO AMMETER EXPERIMENT (CLASS 12)

Aim: To convert a galvanometer of known resistance and figure of merit into an ammeter to read, say 0.05 ampere per division and to calibrate it.

Apparatus: A sensitive Weston (pointer type) galvanometer, a lead accumulator, a rheostat (100Ω), a uniform copper wire of known resistance per unit length, ammeter, key etc.

Theory: To convert a galvanometer into an ammeter to measure any desired value of current, a suitable shunt S is connected to the galvanometer. (A shunt is a low resistance connected in parallel across the terminals of the galvanometer).

If Ig is the figure of merit (current sensitiveness) of the galvanometer, i.e.,the current required to produce a deflection of one scale division, the shunt S required to convert the galvanometer to read i A per division is given by,

ig x G = (i - ig) x S;
Therefore, S = ig x G/(i - ig)
(The shunt S will be a low resistance. Such a low resistance can be obtained from uniform copper wire).

Procedure:

(i) To convert the galvanometer into an ammeter to read 0.05 A/div

The shunt S required to convert the galvanometer into ammeter to read i = 0.05 A/div is calculated using the equation

S = (ig x G)/(i — ig);

Here ig, the figure of merit and G, the resistance of the galvanometer are given. If p is the resistance per unit length of the uniform copper wire, length l required for the shunt is calculated. (l = S/ρ).
The copper wire of length l is cut and connected across the terminals of the galvanometer. Now, the galvanometer is converted into an ammeter to read 0.05 A per division.

(ii) To calibrate the `galvanometer-converted' ammeter

A lead accumulator, a rheostat of resistance about 100Ω, the converted galvanometer, a standard sensitive ammeter (0.1 A/div) and a key are connected in series. The circuit is closed and the rheostat is adjusted to get a deflection of, say, 4 division is the galvanometer. The galvanometer reading is Io = 0.05 x 4 = 0.2 A. The standard ammeter reading 1 is noted. The correction in the converted galvanometer reading, I — Io, is calculated. (The correction may be the positive or negative).

The experiment is repeated for the galvanometer deflections 7, 8, 10, ... (i.e.,for 0.3, 0.4, 0.5, ... A). A calibration curve is drawn with (I — Io) along the y-axis and Io along the x-axis. (Here the points on the graph are joined by straight lines.)

(i) To convert the galvanometer into an ammeter to read 0.05 A/div

Resistance of the galvanometer = ….. Ω

Figure of merit of the galvanometer, ig = ….. A/div

Resistance per unit length of the uniform copper wire, ρ = ….. Ω

All the above quantities are given). Shunt S required to convert the galvanometer into ammeter to read i = 0.05 A/div.

S = (ig x G)/(i - ig) = ..... Ω

Length of the copper wire required for the shunt

l = S/ρ = ..... Ω/cm

(ii) To calibrate the converted galvanometer

i = 0.05 A/div

 Deflection in galvanometer Converted ammeter reading Io(A) Standard ammeter reading I(A) Correction I – Io(A) 4 0.2 6 0.3 8 0.4 10 0.5 - - - -

Result

(i) The galvanometer is converted into an ammeter to read 0.05 A/div.
(ii) The calibration curve is drawn.

Conversion of Galvanometer into Ammeter Experiment Viva Questions and Answers

1. What is an ammeter?

Ammeter is an instrument used to determine the strength of current.

2. What is meant by shunt?

Shunt can be called as a low resistance which is connected in parallel with galvanometer or ammeter.

3. What are the uses of shunt?

i. Shunt is used to protect the galvanometer from strong current.

ii. Shunt is used for converting a galvanometer into an ammeter.

iii. Shunt is used to increase the range of ammeter.

4. Which one has more resistance, galvanometer or ammeter?

Galvanometer

5. Why a galvanometer is having more resistance than an ammeter?

Galvanometer has more resistance than ammeter because ammeter is a shunted galvanometer.

6. Why is the coil wrapped on a conducting frame in a galvanometer?

To make the galvanometer dead beat.

7. Why is the coil of a moving coil galvanometer wound on an aluminium frame?

Because aluminium being a very light metal, does not add significantly to the inertia of the moving coil, also aluminium induces eddy current.

8. Why is soft iron core used in a moving coil galvanometer?

To make the magnetic field radial and to increase the strength of the magnetic field.

9. Why are pole pieces of galvanometer made concave?

Concave poles produce strong, consistent and radial magnetic field.

## Figure of Merit of Galvanometer Experiment (Class 12)

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Figure of Merit of Galvanometer Experiment with Readings (Class 12)

Aim:
To determine the resistance of a pointer type (Weston) galvanometer by half deflection method and find its figure of merit.

Apparatus: The pointer type galvanometer, an accumulator, two resistance boxes Q and R each of range 1 to 1000Ω, a resistance box (P) of low resistance (1 to 10Ω range), rheostat, commutator etc.

Theory:

The current through the resistors P and Q connected in series with the accumulator of emf E is
I = E/(P + Q)

Therefore, P.d across the resistance P = [E/(P + Q)]P = EP/(P + Q)
Therefore, Current through the galvanometer = [EP/(P + Q)] x 1/G,

If d(divisions) is the deflection in the galvanometer, the figure of merit of the galvanometer,

ig = [EP/(P + Q)] x 1/G x 1/d = [E/(P + Q)] x 1/G x P/d A/div:

Procedure
Connection are made as shown in the figure. A resistance 1 Q is introduced in P and a resistance 999 Q is introduced in Q. (The values of P and Q are so adjusted that there is appreciate deflection in the galvanometer.)

The circuit is closed and the deflection d (number of divisions on the scale) of the gal-vanometer is noted without introduced any resistance in R (i.e., R = 0). Now, the resistance is introduced in R to reduce the deflection half of the initial value. The resistance introduced in R gives the resistance G of the galvanometer. The commutator is reversed and the experiment is repeated. The mean deflection d and the mean resistance G are calculated. The value (P/d) is calculated.

The whole experiment is repeated for different values of P keeping (P + Q) constant (i.e., 1000 Ω). The mean values of (P/d) and G are calculated. The figure of merit of the galvanometer is calculated using the equation

ig = i/d = [EP/(P + Q)] x 1/G x 1/d = [E/(P + Q)] x 1/G x P/d

EMF of the accumulator = E = 2V

P+Q = 1000 Ω

 P Ω Q Ω Deflection in galvanometer P/d Resistance in R for half deflection Left Right Mean Left Right Mean 1 999 2 998 3 997 4 996

Mean value of P/d = ….. Mean G = ….. Ω

ig = E/(P+Q) x 1/G x P/d = ….. A/div

Result

(i) The resistance of the galvanometer, G.
(ii) Figure of merit of the galvanometer .

Viva Questions of the figure of merit of a galvanometer Experiment

(i) Give the equation for measuring current using a moving coil galvanometer.

(ii) What is meant by figure of merit?

(iii) What is voltage sensitiveness of a moving coil galvanometer?

(iv) What are the advantages of moving coil galvanometer over the tangent galvanometer.

(v) How can you make a galvanometer sensitive?

(vi) What type of galvanometer is used in the laboratory?

Ans: Weston type (or pointed type)

(vii) How are current sensitivity and figure of merit related to each other.

Ans: Current sensitivity (Si) and figure of merit are inversely proportional to each other.

(viii). What is a galvanometer?

Ans: Galvanometer is an electrical instrument used to study very small current.

(ix). What are different types of galvanometer

Ans: i. Moving coil galvanometer

Eg: Western galvanometer

ii. Moving magnet type galvanometer

Eg: Tangent galvanometer

(x). Which of the above galvanometer is better? Why?

Ans: Moving coil galvanometer is better. Because,

i. It is a dead beat galvanometer.

ii. It can be used in any position.

iii. It has a linear scale.

iv. The external magnetic field does not disturb the working of the galvanometer.

(xi). Why the tangent galvanometer named so?

Ans: Tangent law is the basic working principle of tangent galvanometer.

(xii). Why does the presence of an ammeter near a T.G. affect the deflection of the T.G?

Ans: The magnet in the ammeter produces deflection of the compass needle of the T.G.

(xiii). Why a freely suspended magnet always points along north-south direction?

Ans: Earth is a huge magnet with its N-pole situated near geographic south and S-pole near the geographic north. So a freely suspended magnet always points along north-south direction due to the force of attraction of the opposite poles of earth's magnet.

## Potentiometer Viva Questions with Answers

Potentiometer Experiment Viva Questions with Answers
(i) What is a potentiometer?

Ans: It is an instrument to measure potential difference across two points accurately

(ii) What is its principle?

Ans: A uniform long resistances wire is mounted on a wooden board. When a steady emf is applied to the ends of this wire the fall of potential along the wire is proportional to the length

(iii) How can you increase the accuracy of the potentiometer?

Ans: It can be increased by increasing the length of the wire

(iv) A potentiometer is a better instrument than a voltmeter for measuring emf of a cell. why?

Ans: When we balance a cell against a potentiometer wire no current flows through the cell. Hence we measure the emf. When we use a voltmeter across a cell, a small current flow through the voltmeter and we are getting only the terminal potential difference of the cell

(v) What is meant by potential gradient?

Ans: It is the fall of potential per unit length (mm, cm or m) of the wire when a steady current flows though it.

(vi) What are the parts of a Daniel cell?

Ans: Zinc rod which is the negative plate is placed in a porous pot containing dil: H2SO4. Copper vessel acts as positive plate. The vessel contains copper sulphate solution.

(vii) Why don't you use Daniel cell in the primary circuit of a potentiometer?

Ans: The voltage is small and not steady. It cannot be used for a long time. It is not rechargeable.

(viii) What is the internal resistance of a primary cell due to?

Ans: It is the resistance of the solution. It depends on the nature of the solution, distance between the plates etc

(ix) The terminal potential of a battery while in use is smaller than the emf of the cell. Why?

Ans: Due to internal resistance of the cell there is a potential drop or 'lost volt' within the battery

(x) What is the function of the rheostat in the primary circuit?

Ans: To vary the current in the primary circuit so that we can repeat the experiment

(xi) Aim of the Potentiometer Experiment.

Ans:
(a) To compare the emfs of two cells.
(b) To determine the internal resistance of the given cell using a potentiometer

(xii) Apparatus of the Potentiometer Experiment

Ans: Potentiometer, cells, resistance box, rheostat, galvanometer, two way key, single key, high resistance H R etc.

(xiii) Theory to compare the emfs of two cells

Ans: If E1 and E2 are the emfs of two cells, E1 / E2 = l1 / l2; where l1 and l2 are the balancing lengths of the potentiometer wire for the cells.

(xiv) Procedure to compare the emfs of two cells.

Ans: An accumulator, a rheostat and a key are connected in series with the potentiometer with the positive of the accumulator connected to A. The cells whose e.m.fs are to be compared are connected to a two way key. The common terminals of the two way key are connected to A and to the jockey through a galvanometer and a high resistance (H R) as shown in the diagram. The first cell of e.m.f E1 is included in the circuit by suitably putting the two-way key and the balancing length l1 is measured. Without altering the rheostat in the primary circuit, the cell E2 is included in the circuit and the balancing length l2 is measured.
The emfs of the cell is compared using the equation, E1 / E2 = l1 / l2. The experiment is repeated for different positions of the rheostat and the mean value of E1 / E2 is calculated.

## Post Office Box Experiment Viva Questions

Post Office Box Experiment Viva Questions with Answers

(i) What is the principle of Post Office Box?

Ans: Wheatstone principle P/Q = R/S

(ii) What is the wire used in the resistance box?

Ans: Manganin or constantan

(iii) Does current flow through the galvanometer at the null point?

Ans: The current through the galvanometer is zero

(iv) Why do you press battery key first and then galvanometer key?

Ans: If you press the galvanometer first and then battery key there is induced current in the circuit and is difficult to adjust the balancing

(v) What is temperature coefficient of resistance?

Ans: Temperature coefficient, α = Rt – R0 / R0 x t

(vi) What is meant by negative temperature coefficient?

Ans: The resistance decreases with increase of temperature. Semiconductors have negative temperature coefficient

(vii) Aim of the Post Office Box Experiment?

Ans:
(a)To determine the resistivity of the material of a wire
(b)To verify the laws of resistances

(viii) Apparatus of the Post Office Box Experiment?

Ans: Post office box, resistance wire, battery, galvanometer, battery etc.

(ix)Theory of the Post Office Box Experiment?

Ans: For a balanced Wheatstone's bridge,

P/ Q = R/ X; where P and Q are the resistances in the ratio arms, R is the resistance in the resistance arm and X is the resistor of unknown resistance. If r is the radius and l is the length of X, its resistivity,

ρ = X(πr2/l)

(x) Procedure for determining the resistivity of the material of a wire

Ans: The electrical connection are made as in the figure. Resistances of 10 ohm each are intro-duced in the ratio arms P and Q. Keeping the resistance in the resistance arm R zero, the tap keys K1 and K2 are closed. The direction of deflection is noted. The infinity plug in the resistance arm is now removed. The keys are again pressed. If the deflections are in the opposite directions, the connections are correct. The infinity plug is now inserted. Resistances are introduced in the arm R till the galvanometer shows opposite deflections for two resistances differing by one ohm. Let these resistances be 2 and 3 ohm. According to Wheatstone' principle,

P/Q = R/X = 1

Therefore, X = R. Hence X also lies between 2 and 3 ohm.

Then 100 ohms resistance is taken in the arm P so that the ratio P/ Q = 10 = R/ X;

Therefore, X = R/ 10. As before two resistances differing by one ohm are taken in R so as to get opposite deflections. Let them be 24 and 25 ohm. So X lies between 2.4 and 2.5 ohm. Then 1000 ohms is introduced in R so that P/Q = 100 = R/X;

Therefore, X = R/100.

If 245 and 246 ohms are the resistances to be taken in R so as to get opposite deflections, then X lies between 2.45 and 2.46 ohm. Usually the mean resistances 2.455 ohm is taken as the value of X. The length L and the radius r of the wire are measured. Resistivity is calculated using the formula,
ρ = Xπr2/L

(xi) Procedure for verifying the laws of resistances

Ans:
(a) Resistances in series : The two wires are joined in series and the resistance X3 of the combination is determined. It is found that,
X3 = X1 + X2; (approximately)

(b) Resistance in parallel : The two wires are joined in parallel and the resistance X4 of the combination is determined. It is found that,
X4 =   X1X2/ X1+X2 ; (approximately). Hence Verified.

## Meter Bridge Viva Questions with Answers

Viva Questions and Answers for Meter Bridge Experiment
(i) Why is Wheatstone Bridge so called?

Ans: The principle was discovered by Sir Charles Wheatstone

(ii) What is a Metre Bridge. Why is it so called?

Ans: It is an instrument for comparing resistances. The bridge wire has a length of one metre

(iii) What is the principle of Wheatstone Bridge?

Ans: When the bridge is balanced P/Q = R/S

(iv) Can you find very high resistances accurately with the help of a metre bridge?

Ans: No

(v) Why is the galvanometer is graduated both sides of zero?

Ans: If the jockey is pressed on either side of balance point the deflection in the galvanometer are in opposite directions

(vi) Can we use an ammeter instead of a galvanometer?

Ans: Ammeter is not sensitive compared with a galvanometer. Moreover it is graduated with zero at one end

(vii) What is the material of the bridge wire: What is the criterion for selections?

Ans: Manganin wire is used. Low temperature coefficient

(viii) Why are copper strips on the bridge thick?

Ans: To minimise resistance

(ix) Why is it desirable to have null point as near the centre as possible?

Ans: This will minimise end corrections. Interchanging the resistance also minimises end correction.

(x) When is Wheatstone Bridge most sensitive?

Ans: It is most sensitive when P, Q, R and S are of the same order of magnitude

(xi) What are the laws of combination of resistances?

Ans: R = R1 + R2 for series connection and 1/R = 1/R1 + 1/R2 for parallel connection.

(xii). What is the principle of a metre bridge?

Ans: Wheatstone's principle.

(xiii). What is the material of the wire of metre bridge? Can we use copper wire?

Ans: The wire of metre bridge is generally made of an alloy constantan or eureka. The resistivity of copper is very low and hence it cannot be used in metre bridge.

(xiv). Why do we prefer the null point in the metre bridge experiment in the middle of bridge wire?

Ans: When the null point is in the middle, the ratio arms are nearly equal and the end resistances produce negligible effect.

(xv). Why is it essential that bridge wire should be of uniform area of cross section?

Ans: In deriving the formula, we assume that resistance is proportional to length of the wire. This is true only when the wire is of uniform cross sectional area throughout.

(xvi). Why is metre bridge so called?

Ans: Because bridge uses one metre long wire.

(xvii). When is a Wheatstone's bridge said to be balanced?

Ans: A Wheatstone's bridge is said to be balanced, when the ratio of resistances in arms balance or become equal.

(xviii). What are the applied forms of a Wheatstone's bridge?

Ans: i. Post Office Box

ii. Metre Bridge

(xix). Define electrical conductivity.

Ans: The reciprocal of resistivity is called electrical conductivity.

(xx). What is null point?

Ans: Null point is a point on the wire, keeping jockey at which the galvanometer gives no deflection.

(xxi). How are three resistances connected to get a maximum effective resistance?

Ans: They are connected in series.

(xxii). How does resistance change in series combination?

Ans: Resistance increases.

(xxiii). How does resistance change in parallel combination?

Ans: Resistance decreases.

(xxiv). Explain increase of resistance in series combination.

Ans: In series combination, the effective length of the resistor increases. So resistance increases.

(xxv). Explain decrease of resistance in parallel combination.

Ans: In parallel combination, the effective area of cross section of the resistor increases. So resistance decreases.

(xxvi). What is the effective resistance of two equal resistance connected in parallel?

Ans: Half of a resistance

(xxvii). What is the effective resistance of two equal resistance connected in series?

Ans: Twice of a resistance

(xxviii) Aim of Meter Bridge Experiment

Ans:
(a)To determine the resistivity of the material of a wire
(b) To verify the laws of resistances.

(xxix) Apparatus of Meter Bridge Experiment

Ans: The metre bridge, resistance box, resistance wire, battery, high resistance (HR), key etc.
A metre bridge contains a uniform resistance wire, AB having 1 m length, kept extended between two copper strips set on a wooden board. Another copper strip set centrally on the board leaves two gaps amid the strips. The copper strips are provided with terminals.

(xxx) Theory of Meter Bridge Experiment to determine the resistivity

Ans: If l is the balancing length of the bridge wire from the side of the unknown resistance X and R is the known resistance,

X/R = l/(100 —l);

Therefore, X = R [l/(100 —l)]

Resistivity of X is ρ = X x (πr2 / l); where r is the radius and l is the length of the wire X.

(xxxi) Procedure of Meter Bridge Experiment to determine the resistivity of the material of a wire.

Ans: The unknown resistance X is connected in the left gap and a resistance box R in the right gap. A battery is connected between A and B. A galvanometer is connected between C and a jockey that can slide along the bridge wire.

A suitable resistance R is taken in the box and circuit is closed. The position of the jockey is adjusted so that the galvanometer shows null deflection.

If l1 is the length of balance measured from the left end A, then,
X /R = l1/(100 — l1)

R and X are interchanged in gaps and the balancing length l2 from the right end B is measured. Then,
X/R = l2/(100 — l2)

If l = (l1 +l2)/2;
then X/R = l/(100 — l)
Therefore, X— R[l/(100 — l)]

The experiment is repeated for different values of R; but the balancing point must be near the middle of the bridge wire. In each case X is calculated. From these, the mean value of X is calculated. The radius r of the wire is measured using screw gauge. The length `L' of the wire is measured by a metre scale. Resistivity of the wire is calculated using the equation,

ρ = X πr2/L