Lab Manual Applied Physics PDF

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INSTITUTE OF INDUSTRIAL ELECTRONICS ENGINEERING, (PCSIR) KARACHI Lab. Manual MS113: Applied Physics First Year Spring 2019 Submitted By Submitted To Liaquat Ali Senior Instructor 1 INDEX Lab.No Title Pg. No 1. To Identify the color code of Resistor Band. 3-4 2. To study the characteristics of an acc...

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INSTITUTE OF INDUSTRIAL ELECTRONICS ENGINEERING, (PCSIR) KARACHI

Lab. Manual MS113: Applied Physics First Year Spring 2019 Submitted By

Submitted To Liaquat Ali Senior Instructor

1

INDEX Lab.No

Title

Pg. No

1.

To Identify the color code of Resistor Band.

3-4

2.

To study the characteristics of an acceptor circuit and determine unknown inductance.

5-7

3.

To study the characteristics of a rejecter circuit and determine unknown inductance.

8-9

4.

To determine the Horizontal component of Earth magnetic field strength “H” by Tangent Galvanometer.

10-11

5.

To determine the Capacitance of unknown Capacitor.

12-13

6.

To study the spectral characteristics of photocell and determine the Plank’s constant.

14-16

7.

To determine the Mechanical equivalent of heat “J” by Calendar and Barne’s apparatus

17-18

8.

To determine the velocity of wave propagation in stretched string by using sonometer.

19-20

9.

Determine the unknown high resistance by Neon Flash Bulb Apparatus.

21-22

10.

To determine the charge to mass ratio (e/m) of Electron (Lorentz Force).

23-25

11.

Speed Measurement of motor using Strobe Light.

26

2

Applied Physics

Color coding

IIEE

1st Semester

Institute of Industrial Electronics Engineering

LAB 1

OBJECTIVE: To Study and Identification of Resisters, Capacitors by using Color Codes and Capacitors by coding. THEORY:

Resistor Color

Code Guide Capacitor Letter Codes Table Picofarad Nanofarad Microfarad (pF) (nF) (uF)

Code

Picofarad Nanofarad Microfarad (pF) (nF) (uF)

Code

10

0.01

0.00001

100

4700

4.7

0.0047

472

15

0.015

0.000015

150

5000

5.0

0.005

502

22

0.022

0.000022

220

5600

5.6

0.0056

562

33

0.033

0.000033

330

6800

6.8

0.0068

682

3

47

0.047

0.000047

470

10000

10

0.01

103

100

0.1

0.0001

101

15000

15

0.015

153

Fig 3.4.1 Four Band Standard EIA Colour Code For Inductors.

Values are in micro Henries (μH) First two digits are the value Third digit is the multiplier If there is an R, its acts as a decimal point, and there is no multiplier Examples: 101 = 10*101μH = 100μH 4R7 = 4.7μH Suffix Sometimes the precision of the inductor will be marked, using a final letter F, G, J, K, or M F = +/-1% G = +/-2% J = +/-5% K = +/-10% M = +/-20%

Band

1

2

Meaning 1st Digit 2nd Digit

3

4

Multiplier (No. of zeros)

Tolerance %

Gold

x 0.1 (divide by 10)

+/-5%

Silver

x 0.01 (divide by 100)

+/-10% +/-20%

Black

0

0

x1 (No Zeros)

Brown

1

1

x10 (0)

Red

2

2

x100 (00)

Orange

3

3

x1000 (000)

Yellow

4

4

x10000 (0,000)

Green

5

5

Blue

6

6

Violet

7

7

Grey

8

8

White

9

9

4

Applied Physics 1st Semester

Acceptor circuit

IIEE

Institute of Industrial Electronics Engineering

LAB 2

OBJECTIVES: To study the characteristics of an acceptor circuit and determine unknown inductance. APPARATUS: Resistance, capacitors, inductor, a frequency generator/oscillator, an Oscilloscope, connecting wires, A.C. supply etc. THEORY: Acceptor Circuit Consider a circuit consisting of a resistor R, an inductor L, a capacitor all connected in series in an A.C. circuit. Then e.m.f. applied across such a circuit has to overcome opposition offered by all the above three circuit elements. Let I be the current flowing in the circuit at any instant then: -------- (1) Where E is the applied voltage and is given by E = Eo sinωt, ω is angular frequency = 2πf , and f is linear frequency of the applied voltage If Io be the max. current that flows through the circuit then solving equation (1) we will get:

= 1 is the total reactance, C is the total impedance to the flow

Where X =  L  Z= of A.C. Hence

Io 

Eo Z

---------- (2)

From eq.(2) we see that the amplitude of the current can assume maximum value if the impedance becomes minimum and this becomes feasible. 1   Where the term  L   becomes zero for a C   certain value of angular frequency ω. Consider the following two extreme cases: 1   (i). For ω = 0  L   =  therefore Io= 0 C  

Thus the current Io is zero, both for very low and very high frequencies. For any other value of angular frequency ω lying in the range 0 < ω <  the value of the current depends on the reactance 1    L   . If the angular frequency is increased C   then the inductive reactance increases till the angular frequency has such a value that the inductive reactance is equal to capacitive reactance 1   i.e.  L   at this stage the current has max. C   value. Hence the condition for maximum current to flow in the circuit is 1 1 ωL = or ω2 = C LC 1 but ω = 2πf ω= LC 1 f= --------- (3) 2 LC This frequency is called the resonant frequency. We see that at resonate frequency inductive reactance cancels the capacitive reactance and the current are then entirely determined by the resistive element R of the circuit. Hence at resonance the maximum current is given by: Io=Eo /R At this frequency the impedance is minimum and it is equal to R and the current is in phase with the applied voltage. If the frequency of the applied voltage is further increased beyond the resonant frequency, than the inductive reactance will increase and so the current will go on decreasing. Such a circuit is called a series resonant or most commonly an acceptor circuit.

5

1   (ii). For ω =   L   = + therefore Io= 0 C  

WORKING FORMULA: Inductance is given as

L

1 4 2 f 2 c

PROCEDURE: 1. Make the circuit connections as shown in circuit diagram i.e. take a resistor, an inductor and a capacitor of suitable values connect them with each other in series and connect them across a frequency generator and cathode ray oscilloscope (CRO). 2. Apply a signal of certain voltage by the oscillator at a certain frequency say 100 kHz and note down the amplitude of output signal from CRO. 3. Now start increasing frequency of input signal (keeping the voltage constant), step by step, each time note down the amplitude of the output signal, we will see that amplitude of the output signal will first increase steadily reaches a maximum value and then start decreasing. 4. Plot a graph b/w frequency (f) on x-axis and the corresponding amplitude of the signal on yaxis for each set of observation on one graph paper. 5. From graph calculate resonant frequencies for capacitor at which amplitude of signal is maximum. 6. Calculate the inductance with the help of resonant frequencies obtained through graph. OBSERVATIONS: Capacitance of capacitor C =------------µF S. Frequency Amplitude S. Frequency Amplitude No. (f)Hz (f)Hz of Signal No. of Signal division division 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 CIRCUIT DIAGRAM:

6

CALCULATIONS:

From Graph: Value of resonant frequency f = --------Hz; 1 L 2 2 4 f c

RESULT: 1. Characteristics of an acceptor circuit studied. it is seen that at resonant frequency the output signal is maximum/minimum. 2. The value of inductance is found L = ----------- Henry.

PRECAUTIONS AND SOURCES OF ERROR: 1. Before switching on, get connections checked by teacher. 2. The amplitude of the signal at resonant frequency may not be exactly zero, which is due to presence of resistance. 3. Choose proper combinations of L, R & C.

7

Applied Physics

Rejecter Circuit

IIEE

1st Semester

Institute of Industrial Electronics Engineering

LAB 3

OBJECTIVES: To study the characteristics of a rejecter circuit and determine unknown inductance. APPARATUS: Resistance, capacitors, inductor, an oscillator, an oscilloscope, connecting wires, A.C. supply etc. THEORY: REJECTER CIRCUIT If an inductor and a capacitor are connected in parallel and an A.C. voltage is applied across them, then in this case it is the voltage rather than current, which is the same on each element of the circuit. The current is at resonance is minimum. This is so because the current in the capacitive branch is in opposite phase to the current in the inductive branch. Since the current is rejected by the parallel combination of LC at resonance hence it is termed as the rejecter circuit. In this case, the impedance is at maximum. If there is no resistance in the circuit Then the resonant frequency is given by: 1 f= 2 LC and the current is given by E Io  o Z Where Z is the total impedance of the circuit. Since at resonance, impedance is infinite, so the current almost reduces to zero.

WORKING FORMULA:

Inductance is given as

L

1 4 2 f 2 c

CIRCUIT DIAGRAM:

PROCEDURE: 7. Make the circuit connections as shown in circuit diagram i.e. take a resistor, an inductor and a capacitor of suitable values connect them with each other in parallel and connect an oscillator and cathode ray oscilloscope (CRO) across this combination of LRC. 8. Apply a signal of certain voltage by the oscillator at a certain frequency and note down the amplitude of output signal. 9. Now start increasing frequency of input signal (keeping the voltage constant), step by step, and each time note down the amplitude of the output signal. Take a number of observations. We will see that at first the amplitude of the output signal decreases as the frequency increases; it finally attains its minimum value and then again rises. At the resonant frequency the amplitude is minimum. 10. Plot a graph b/w frequency (f) on x-axis and the corresponding amplitude (A) of the signal on y-axis. 11. Calculate the inductance with the help of resonant frequencies obtained through graph. 8

OBSERVATIONS: Capacitance of capacitor C =------------µF S. Frequency Amplitude S. Frequency of Signal No. (f) KHz No. (f) KHz division 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 CALCULATIONS:

Amplitude of Signal division

From Graph:

Value of resonant frequency f = --------Hz; 1 L 2 2 4 f c RESULT: 1. Characteristics of an rejecter circuit studied. it is seen that at resonant frequency the output signal is minimum/maximum. 2. The value of inductance is found L = ----------- Henry. PRECAUTIONS AND SOURCES OF ERROR: 1. Before switching on, get connections checked by teacher 2. The amplitude of the signal at resonant frequency may not be exactly zero, which is due to presence of resistance. 3. Choose proper combinations of L, R & C.

9

Applied Physics

Tangent Galvanometer

IIEE

1st Semester

Institute of Industrial Electronics Engineering

LAB 4

OBJECTIVES: To determine the horizontal component of Earth Magnetic field strength “He” by Tangent Galvanometer. APPARATUS: Tangent galvanometer, coil, reversing key, connecting wires and ammeter. THEORY: Two magnetic fields can be compared by means of tangent galvanometer, which consist of a magnet pivoted on a vertical axis and carrying a light pointer which can move over a circular scale. Normally one of the fields is earth horizontal component and other field is arranged to be at right angle to this. The pivoted magnet sets itself along the resultant of the two fields at an angle ϴ to its direction when it is in the earth magnetic field alone. If He is a magnetizing force of the earth horizontal component and H is the magnetizing force of the other field (coil). The magnetizing force at the center of a circular coil of known radius and known number of turns, when a measured current is passing through the coil, can be calculated. This can be compared by means of a tangent galvanometer, with the horizontal component of the earth’s magnetizing force, enabling the other to be determined. If n is the number of turns in the coil, r meter is its radius, 1 amp is the current, and H amp per meter the magnetizing force at the center of the coil. H=nI/2r amp per meter Using equation (1) H= He tan He=H/ tan He= NI / 2r tan amp per meter 1 amp. per meter = 0.01256 oersted.

CIRCUIT DIAGRAM:

PROCEDURE: 1. Connect the circuit as shown in figure. 2. Using leveling screw adjust dial of magnetometer, so that needle could not strike the dial. 3. Adjust the needle of tangent galvanometer so that, the needle is perpendicular to the circular coil. 4. Set rheostat for minimum current (i.e. Max resistance), turn on the circuit and give 100,150, 200, 250 and 300 clockwise deflections to magnetometer needle by increasing the current with rheostat. 5. Reverse the keys of reversing switch and repeat the step 4 in anticlockwise direction 6. Note down the corresponding current for each deflection. 7. Plot a graph between I and tan ϴ, which will be a straight line of slope I/tan ϴ. 8. Knowing also N and r, He can be calculated .

10

OBSERVATIONS: No. of turns of a coil = n = ____________ turns Radius of the coil = r = _____________m Current in mA S.No.

Angle  (deg)

Clock Wise Reading I1 (mA)

Anti-Clock Wise Reading I2 (mA)

MEAN I= (I1+I2)/2 (mA)

tan ϴ

01 02 03 04 05 CALCULATION: NI He = amp per meter 2r tan  1 amp per meter = 0.01256 oersted.

RESULT: The value of He is found to be =_____________ (oersted) PRECAUTIONS: 1. Connections should be tight. 2. Angle should be adjusted with care. 3. Dial of galvanometer should be leveled properly. 4. Keep metallic material away from magnetic dial.

11

Applied Physics

Capacitance of unknown Capacitor

IIEE

1st Semester

Institute of Industrial Electronics Engineering

LAB 5

OBJECTIVES: To determine the capacitance of an unknown capacitor and To study the charging and discharging of a capacitor. EQUIPMENT 22 kΩ resistor, 470 f capacitor, voltmeter, stopwatch, toggle switch, bread board, Power supply THEORY

CIRCUIT DIAGRAM

CHARGING Consider the circuit diagram shown, the capacitor charging equation is given by

Where

= supply voltage.

At specific value of time, t = RC, called the time constant of the RC circuit is

Fig 1 Charging a capacitor

Therefore, by plotting Vc versus t, the time constant, τ, can be determined, and hence, the value of C, if R is known. Fig 2 Discharging a capacitor For the discharging process, consider the circuit shown in Fig. 2. When the switch, S, is closed, the capacitor is charged up to a value of Q=Cε. As the switch is opened, the electromotive force is disconnected from the circuit, and the capacitor starts to discharge through the resistor. The discharging equation is given by :

The voltage across the capacitor, Vc, is derived to be At τ =RC

Therefore, by plotting Vc versus t, the time constant τ can be determined, and hence the value of capacitor if R is known.

PROCEDURE: CHARGING 1. Connect the circuit as shown in Fig. 1 (make sure that the lead of the capacitor at the arrow head side, is connected to the ground). 2. Turn on the power supply. Set the output of the power supply to 10 V. 3 Short out the capacitor, temporarily, by connecting a wire parallel to it (so that the capacitor is completely discharged). 4 Close the switch, S, and reset the stopwatch. 5 Simultaneously, remove the shorting wire, and start the stopwatch. 6 The capacitor will start charging up, and the voltage across the capacitor, Vc, will increase correspondingly.

12

PROCEDURE: CHARGING

Cont’d 7 Corresponding to integer values of Vc , from (1-8) V by 1 V increments, stop the stopwatch, temporarily, record the time, t, in Table I under t1 heading, and restart the stopwatch. 8 Stop the stopwatch. 9 Repeat steps 3-8. Record the measured time under t2. 10 Calculate the average time, tavg. 11 Plot a graph for Vc versus tavg, from which, determine τ, and calculate C.

PROCEDUR: DISCHARGING 1 Connect the circuit as shown in Fig.2 (make sure that the lead of the capacitor at the arrow head side, is connected to the ground). 2 Turn on the power supply, and set the voltage to10 V. 3 Close the switch, S. This will cause the capacitor to charge up immediately. 4 Start the stopwatch and open the switch, S, simultaneously. Table-1 (CHARGING) Vc(V) 0 1 2 3 4 5 6 Table-2 (DISCHARGING) Vc(V) 10 9 8 7 6 5 4 3

T1(Sec)

T1(Sec)

5

6

7 8 9 10 11

12 13

The capacitor will start discharging through the resistor, R, and the voltage across the capacitor, Vc, will decrease correspondingly. Corresponding to integer values of Vc, according to Table II, stop the stopwatch, temporarily, record the time, t, in Table II under t1 heading, and restart the stopwatch Stop the stopwatch. Repeat steps 3-7. Record the measured time under t2. Calculate the average time, tavg. Determine ln(Vc) for all the values, using a calculator. Plot Vc versus tavg on the same graph of part one, from which, determine τ, and calculate C. Plot another graph for ln(Vc) versus tavg, from which, determine ε and C. Calculate the average value of C, using the values obtained from parts one & two.

T2(Sec)

T2(Sec)

Tavg(Sec)

Tavg(Sec)

lnVc

13

Applied Physics

Plank’s constant

IIEE

1st Semester

Institute of Industrial Electronics Engineering

LAB 6

OBJECTIVES: To study the spectral characteristics of photovoltaic cell and determine the Plank’s Constant and work function. APPARATUS: Planks constant measuring Instrument. THEORY: When light falls on certain material like selenium etc. (which has low work function). Electrons are emitted from their surfaces. These electrons are called photoelectric and the phenomenon governing such an emission is called the photoelectric effect, the material is said to be photosensitive material. The energy quantization of electromagnetic radiation in general, and of light in particular, is expressed in the famous relation

E=hf

(1)

where E is the energy of the radiation, f is its frequency, and h is Planck's constant (6...