.

Experiment :C-21. Measurement of the phase difference between VR

and VC

in an RC circuit from Lissajous figures.

2. Study of the frequency response of

(a) Phase shift between Vin

and VC

and, (b) Peak charge value of the capacitor in an RC circuit.

Submitted to Mr.Golam Dastegir Al-Quaderi and

Dr. Ratan Chandra Gosh

Submitted by Muhammed Mehedi Hassan Group A ;Batch-09 Second Year, Roll–SH 236 Student of Physics Department, Uinversity of Dhaka.

Date Date ofexperiment of submission April May 10,2011. 19,2011.1

Experiment :C-21. Measurement of the phase difference between V

R

and V

C

in an RC circuit from Lissajous figures.

2. Study of the frequency response of

(a) Phase shift between V

in

and, (b) Peak charge value of the capacitor in an RC circuit.

Theory : if two simultaneous simple harmonic motions are applied at right angles to each other, a Lissajous figure is obtained. If the signals have the same frequency with a constsnt phase difference between them, the resulting figure is an ellipse in general. It is obvious that the ellipse will be confined within a rectangular space with lengths A

x

and V

C

and A

y

, the amplitudes of the waves. However, the shape of the ellipse and and its orientations depends on the ratio between the amplitudes as well as phase difference δ. Patterns for different typical values of δ are shown below.

Figure 2.1 Shape of Lissajous figures for various phase differences

Let the two waves be,

x = A

x

cos ωt (1)

y = A

y

cos(ωt + δ) (2)

Equation (1) ⇒ cosωt= A x

x

Equation (2) ⇒

A y

y

=cosωtcosδ−sinωtsinδ=cos ωtcosδ−

√

1 − cos2 ωt sinδ

2

Now equation (1) and (2) can be combined like this:

y A

y

=

A x

x

cos δ −

√

1 − (

A x

x

)2 sinδ (3)

⇒ sin2 δ =

A2 x2

x

+A2 y2

y

−

A

2xy

x

cosδ (4)

The phase angle δ can be obtained from the following considerations.In general the equation (4) represents an inclined ellipse as shown in figure 2.2.

a b

Figure 2.2 .

At x = 0 , sin2 δ =

A

y

y2 A2 y Hence |sinδ|=

A |y|

y

=

2A 2|y|

y

=

PQ RS

= B A

Considering the absolute value,

|δ| = sin−1

B A

(5)

Thus |δ| can be found by measuring A and B [fig 2.2.a] or, A and B [fig 2.2.b]

For a circle or vertical When a sinusoidal voltage ellipse source B=A, applied

and |δ| =sin−1 1 = π 2

V

in

= V cos ωt (6)

In the following circuit:V

in

= V

R

+ V

C

3

Which , in terms of the charge q(t) on the capacitor at time t, becomes

V cos ωt =

dq(t) dt

q(t) C

(7)

It can be shown that

q(t) = Qcos(ωt + φ) (8)

Equation (3) clearly show that V

R

R +

Apparatus :leads V

C

by an angle of π/2

An oscilloscope. signal generator, a resistor, a capacitor and a circuit board.

Figure 2.3 Circuit diagram to determine the phase difference between V

R

& V

C

Figure 2.4 Circuit diagram to determine the phase difference between V

C

& V

in

.

4

& Table-1.Determination of Phase difference between V

R

V

C

:

Frequency

Type of Figure A (Hz) B (div) (div) B A

|δ| Degree = sin−1 B A

Mean|δ| Degree

250Vertical Ellipse

2.1 2.1 1 90

300 ” 2.1 2.1 1 90

400 ” 2.1 2.1 1 90

500 ” 2.3 2.3 1 90

600 ” 2.3 2.3 1 90

700Near Circle

2.3 2.3 1 90

800 ” 2.5 2.5 1 90

900 ” 2.7 2.7 1 90

1000 ” 2.9 2.9 1 90

1100Horizontal Ellipse

1.0 1.0 1 90 90

1200 ” 1.2 1.2 1 90

1300 ” 1.3 1.3 1 90

1400 ” 1.4 1.4 1 90

1500Ellipse

3.4 3.4 1 90

1600 ” 3.5 3.5 1 90

1700 ” 3.7 3.7 1 90

1800 ” 3.8 3.8 1 90

2000 ” 4.0 4.0 1 90

3000 ” 4.5 4.5 1 90

4000 ” 4.7 4.7 1 90

5000 ” 5.0 5.0 1 905

Table -2. Measurement of ∆φ between VC

and V

in Frequency

f lnf A B φ = sin−1 B A

φ ×360 = ∆tT

(Hz) (div) (div) Degree (ms) (ms) Degree 50 3.91 7.80 0.40 2.94 0.10 19.40 0.005 1.86

100 4.61 4.10 0.25 3.49 0.08 9.20 0.008 3.13 150 5.01 5.25 0.40 4.37 0.08 6.40 0.013 4.5

200 5.29 5.30 0.60 6.50 0.10 4.75 0.021 7.58 300 5.70 5.30 1.00 10.88 0.10 3.15 0.032 11.43 400 5.99 5.25 1.30 14.34 0.10 2.48 0.040 14.51 500 6.22 5.30 1.65 17.84 0.10 1.92 0.052 18.75 600 6.40 5.30 1.90 21.01 0.11 1.80 0.061 22.00 700 6.55 5.01 2.20

26.04 0.09 1.40 0.064 23.14 800 6.69 5.05 2.40 28.38 0.09 1.22 0.073 26.55 900 6.80 5.00 2.55 30.66 0.09 1.08 0.083 30.00 1000 6.91 4.85 2.70 33.83 0.08 0.94 0.085 30.63

1300 7.17 4.05 2.55 39.02 0.07 0.77 0.091 32.73 1500 7.31 3.80 2.65 44.22 0.07 0.65 0.108 38.77 1800 7.50 3.45 2.60 48.91 0.06 0.53 0.113 40.75 2000 7.60 3.20 2.55

52.83 0.06 0.47 0.128 45.96 3000 8.01 2.45 2.15 61.35 0.05 0.31 0.161 58.06 4000 8.29 2.05 1.85 64.48 0.04 0.25 0.160 57.60 5000 8.52 1.65 1.55 69.95 0.04 0.20 0.200

72.00 6000 8.70 1.40 1.35 74.64 0.03 0.16 0.188 67.50 7000 8.85 1.15 1.12 76.88 0.03 0.14 0.214 77.14 7500 8.92 4.50 4.45 81.45 0.03 0.135 0.222 80.00 8000 8.99 1.00

1.00 90.00 0.03 0.130 0.232 82.80 9000 9.11 1.00 1.00 90.00 0.02 0.104 0.218 78.48 10000 9.21 0.95 0.95 90.00 0.02 0.09 0.222 80.00 15000 9.62 1.20 1.20 90.00 0.02

0.09 0.222 80.006

∆t T ∆tT

Table -3. Frequency Response of Peak Charge

Frequency lnf Voltage Voltage Voltage Voltage Peak

Charge f in out gain gain

Hz V V |V Vout in

|V | 20 log10 dB out

V

in

| Q = |Vout

| × C Coulomb 30 3.40 5.40 5.40 1 0 5.40 × 10−8 40 3.69 5.30 5.30 1 0 5.30 × 10−8 50 3.91 5.25 5.25 1 0 5.25 × 10−8 60 4.09 5.25 5.25 1 0 5.25 × 10−8 70 4.25 5.20 5.20 1 0 5.20 × 10−8 100 4.61 5.20 5.20 1 0 5.20 × 10−8 200 5.30 5.25 5.20 0.99 -0.080 5.20 ×

10−8 300 5.70 5.25 5.20 0.99 -0.080 5.20 × 10−8 500 6.22 5.40 5.30 0.98 -0.160 5.30 × 10−8 1000 6.91 5.20 4.20 0.81 -01.85 4.20 × 10−8 2000 7.60 5.60 3.18 0.57 -4.88 3.18

× 10−8 3000 8.01 5.80 2.40 0.41 -7.74 2.40 × 10−8 4000 8.29 5.80 2.00 0.35 -9.12 2.00 × 10−8 5000 8.52 6.00 1.70 0.28 -11.06 1.70 × 10−8 6000 8.70 6.00 1.40 0.23 -12.77 1.40 × 10−8 7000 8.85 6.20 1.20 0.19 -14.43 1.20 × 10−8 8000 8.99 6.30 1.10 0.17 -

15.39 1.10 × 10−8 9000 9.11 5.78 0.81 0.14 -17.11 0.80 × 10−8 10000 9.21 5.81 0.50 0.086 -21.30 0.50 × 10−8

7

Calculation : The resistor we used has a resistance of 10 kΩ :

R=10 kΩ=10×103Ω

The capacitance C=0.01µ F=0.01×10−6F From the Graph of φ vs lnf frequency fφ

,

fφ

=elnf = e7.45 = 1.7 × 103Hz

The Cut-Off frequency or the half power frequency ; from the second graph “ln f vs Voltage gain” at voltage gain=0.717V

fg

= elnf = e7.3 = 1.4 × 103Hz

From the third graph“lnf vs 20 log Voltage gain” the Cut-Off fre- quency at -3dB

fg

= elnf = e7.2 = 1.3 × 103Hz

Average gain frequency,

fg

=f

g

+f

g 2

= 1.4×103Hz+1.3×103Hz

2

= 1.35 × 103Hz

Average frequency of fφ

and fg

is,

f =f

φ

+f 2

g

= 1.7×103Hz+1.35×103Hz

2

= 1.53 × 103Hz

Percentage of error in comparison between them∆τ = |1.7−1.35

1.53

| × 100%

=22.80%

Experimentally τ

exp = Theoretical value of τ1 ω

= 1

2πf

= 2×3.1416×1.53×1000

1

s = 9.6ms

τtheoretical

= RC = 10 × 103 × 0.01 × 10−6s=10ms

Percentage of error in comparison between them;

∆τ = |τ

theoretical

τ

theoretical −τ

exp

| × 100%= |10−9.6

10

| × 100%

=4.0%8

Result :1. The phase difference between VR

and VC

is 90 ˙ .

2. The frequency response

(a) From second graph the frequency is 1.4×103 Hz and from

third graph 1.3 × 103 Hz and, (b) Peak charge value of the capacitor is 5.4×10−8 C.

Discussion :Our experiment showed that the capacitor current is 90 out of phase with the

resistor (and source) current.Therefore the first ex- periment was solved successfully. Performing RC oscillator experi- ment we got the “feeling” for the actual phase of the RC oscillator and its relation betweeen them. Doing the experiment we were able to identify the cut-off frequency in two different methods. Our re- sult have some deviation which is not very significant. It was due to some experimental limitations. Firstly, the time period taken from the gridline of the oscilloscope monitor was not accurate. When the frequency was high(above 1200 Hz) it was hard to find the actual di- vision of the gridline during one complete cycle. Secondly, the signal generator has no calibration between 1000Hz to 1100Hz ; we have to assume this unmarked frequency as well as the height of A and length of B, so there may be some error. Due to all these reason and other experimental limitations our resulting cut-off frequency has 22.80% error.9