Exp.Date
17.04.13
Report Score Delay Score
Coming Date
24.04.13
Result Score
KADIR HAS UNIVERSITY
ENGINEERING FACULTY
DEPARTMENT OF
ELECTRONICS ENGINEERING
EE 204 CIRCUIT LABORATORY
EXERIMENT NO:
EXPERIMENT NAME:
18
Part 1 : Impedance of series AC Circuits
Part 2 : Admittance of parallel AC Circuits
GROUP NO: 2
GROUP MEMBERS:
ANIL YEŞİLKAYA
Delay
Signature

OBJECTIVE
In this experiment, we try to measure

Impedance and phase between the AC voltage and current for a series RL circuit

And compare measured values with calculated values

Impedance and phase between the AC voltage and current for a series RC circuit

And Compare measured values with calculated values

Impedance and phase between the AC voltage and current for a series RL circuit

And Compare measured values with calculated values

Admittance and phase difference between AC voltage and current for a parallel RL
Circuit

And Compare measured values with calculated values

Admittance and phase difference between AC voltage and current for a parallel RC
Circuit

And Compare measured values with calculated values

Admittance and phase difference between AC voltage and current for a parallel RLC
Circuit

And Compare measured values with calculated values
COMPONENTS AND EQUIPMENTS

C. A. D. E. T

One function generator

One dual-trace oscilloscope

One 0-10mA ac milliammeter

One 0.1 microFarad capacitor

One 100 miliHenry inductor

One 1k ohm resistor

One 0-20mA ac milliammeter

One 0-10 V ac voltmeter

Resistors 1 ohm and 1k ohm
GENERAL THEORY
General theory is comparing measured and calculated phase difference RLC Circuits and AC voltages.
Oscilloscope is electronic device that test circuits and determine signal at different points. In this experiment, we saw voltage across resistance on the oscillator screen.
Inductor: Inductor is an electrical component that stands out any change of electrical current. Voltage across between the terminals of the inductor can be formulated as ;
V = L x 𝜕𝑖/𝜕𝑡
In addition, if the current is constant , it means that voltage is zero. Secondly , the current can not change instantaneously for inductor.

Capacitor: Capacitor is electrical component that consists of two conductors separated by an insulator. The current flows between capacitor terminals can be formulated as
I = C x 𝜕𝑉/𝜕𝑡
In addition, if the voltage across the capacitor is constant, it means that capacitor current is zero. Secondly, voltage cannot change instantaneously for capacitor.
Ohm’s law is used for this circuit but the only change is impedance instead of resistor.
V = Z x I
We must know some formulas for this experiment, they are shown in below.
Voltage and current values may be peak values for this equation(upside).
Z is impedance of the circuit and it can be formulated as ;
Z = √𝑅 2 + 𝑋 2
X is reactance of the circuit.
We calculated phase difference between AC rms voltage and AC rms current like this; 𝜑 = Arctan ( X / R )
= 𝜑 is phase difference in radians or degrees
Impedance of RL Circuit is ;
Z = √𝑅 2 + (𝑋𝑙) 2
Phase difference ; 𝜑 = Arctan ( (X 𝑙 ) / R )
Impedance of RC Circuit is;
Z = √𝑅 2 + (𝑋𝑐) 2
Phase difference; 𝜑 = (−) Arctan ( (X 𝑐 ) / R )
Impedance of RLC Circuit is ;
Z = √𝑅 2 + (𝑋) 2
Phase difference; 𝜑 = Arctan ( (X) / R ) , X is total reactance = X 𝑙 - X 𝑐
Because inductive reactance and capacitive reactance have 180 degree phase difference.

There are also some formulas we need to know for the second experiment;
Admittance (Y ) = 1 / Z
V = Z x I is equal to Y x V = I (it is valid for peak values)
Parallel circuit admittance is Y = √𝐺 2 + (𝐵) 2
G = 1/ R = Conductance
B = 1/X = Susceptance
Phase degree between AC rms voltage and AC rms current can be formulated ; 𝜑 = Arctan ( B / G )
For RL Circuit;
Y = √𝐺 2 + (𝐵𝑙) 2 𝜑 = (−) Arctan ( Bl / G )
Bl = 1/ Xl = 1/ 2*pi*f*L
For RC Circuit;
Y = √𝐺 2 + (𝐵𝑐) 2 𝜑 = Arctan ( Bc / G )
Bc = 1/ Xc = 2*pi*f*C
B net = Bc – Bl
Because inductive reactance and capacitive reactance have 180 degree phase difference.
For RLC Circuit;
Y = √𝐺 2 + (𝐵) 2 𝜑 = Arctan ( B / G )

PROCEDURE
This experiment consists of 2 parts.
PART 1;
In the first step;
1.
We generated a circuit in figure 18.1.(100 mH and 1k ohm are series)
2.
We found phase difference between voltage and current
3.
12,5 notch => 360 degree 2 notch is 57.6 degree (voltage front)
4.
Voltage is 7.07 V rms , current is 3.53 mA rms
5.
Z = 7.07 V / 3.53 mA = 2k Ohm
6.
Xl = 2*pi*f*L = 2*3.14*2KHz*0.1H = 1256 ohm
7.
Z = √𝑅 2 + (𝑋𝑙) 2
= 1605 ohm
8.
Expected phase difference can be found ;
9.
Xl = 2*pi*f*L = 2*3.14*2KHz*0.1H = 1256 ohm
10.
𝜑 = arctan (1256 /1000) = 51.7 degrees
In the second step ;
11.
We generated a circuit in figure 18.2.(0.1mikroF and 1k ohm are series)
12.
We found phase difference between voltage and current
13.
13 notch = > 360 degree , 1 notch = 27.6 degree (current is front )
14.
Voltage is 7.07 V rms , current is 3.53 mA rms
15.
Z = 7.07 V / 3.53 mA = 2k Ohm
16.
Xc =1/ 2*pi*f*C = 1/2*3.14*2KHz*0.1 mikroF = 796 ohm
17.
Z = √𝑅 2 + (𝑋𝑐) 2
= 1278 ohm
18.
Expected phase difference can be found ;
19.
Xc =1/ 2*pi*f*C = 1/2*3.14*2KHz*0.1 mikroF = 796 ohm
20.
𝜑 = (−)arctan (796 /1000) = - 38 degrees
21.
𝑊𝑒 𝑢𝑠𝑒𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑤𝑒 𝑝𝑢𝑡 − 𝑠𝑖𝑔𝑛 𝑡𝑜 𝑝ℎ𝑎𝑠𝑒 𝑑𝑒𝑔𝑟𝑒𝑒.

In the third step ;
22.
We generated a circuit in figure 18.3.(0.1mikroF , 0.1H and 1k ohm are series)
23.
The only change is frequency become 1KHz.
24.
We found phase difference between voltage and current
25.
10 notch = > 360 degree , 1 notch = 36 degree (current is again front )
26.
Voltage is 7.07 V rms , current is 3.53 mA rms
27.
Z = 7.07 V / 3.53 mA = 2k Ohm
28.
Xc =1/ 2*pi*f*C = 1/2*3.14*1KHz*0.1 mikroF = 1592 ohm
29.
Xl = 2*pi*f*L = 2*3.14*1KHz*0.1H = 628 ohm
30.
X = 628 – 1592 ohm = -964 ohm Capacitive circuit
31.
Z = √𝑅 2 + (𝑋) 2
= 1388 ohm
32.
Phase difference is that ;
33.
X = 628 – 1592 ohm = -964 ohm
34.
= arctan (-964 / 1000 ) = -43.94 degree
35.
Capacitive circuit so that – sign is put the phase angle.

PART 2;
In the first step;
1.
We generated a circuit in figure 18.4.(100 mH and 1k ohm are parallel, 1 ohm are series)
2.
We found phase difference between voltage and current
3.
10 notch => 360 degree 1,5 notch is 54 degree (current back, voltage is front)
4.
Voltage is 7.07 V rms , current is (0.02 V / √2 = 0.014 A rms
5.
Y x V = I , Y = (0.014 mA / 7.07 V ) = 2 mili Siemens
6.
Z = 1 / Y = 500 ohm
7.
Bl = 1/ Xl = 1/ 2*pi*f*L = 1 / 2*3.14*1KHz*0.1 = 1.59 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
8.
G = 1/R = 0.001 ohm
9.
Y = √𝐺 2 + (𝐵𝑙) 2
= 1.87 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
10.
Z = 1/Y = 532 ohm
11.
Bl = 1.59 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠 and G = 1/R = 0.001 ohm
12.
𝜑 = (−) arctan(1.59 𝑥 10 −3 /0.001) =
−57 𝑑𝑒𝑔𝑟𝑒𝑒 (𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑏𝑎𝑐𝑘 𝑎𝑛𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑠 𝑓𝑟𝑜𝑛𝑡) Inductive circuit

In the second step ;
13.
We generated a circuit in figure 18.5.(0.1 mikroF and 1k ohm are parallel, 1 ohm are series) frequency become 2kHZ.
14.
We found phase difference between voltage and current
15.
24 notch => 360 degree 4 notch is 60 degree (voltage back , current is front )
16.
Voltage is 7.07 V rms , current is (0.02 V / √2 = 0.014 A rms
17.
Y x V = I , Y = (0.014 mA / 7.07 V ) = 2 mili Siemens
18.
Z = 1 / Y = 500 ohm
19.
Bc = 1/ Xc = 2*pi*f*C = 2*3.14*2KHz*(0.1 x 10 −6 ) = 1.25 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
20.
G = 1/R = 0.001 ohm
21.
Y = √𝐺 2 + (𝐵𝑐) 2
= 1.60 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
22.
Z = 1/Y = 622 ohm
23.
Bc = 1.25 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠 and G = 1/R = 0.001 ohm
24.
𝜑 = arctan(1.25 x 10 −3 /0.001) = 51 𝑑𝑒𝑔𝑟𝑒𝑒 (current is front , voltage is back)
25.
𝑊𝑒 𝑠𝑒𝑡𝑡𝑒𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑎𝑛𝑑 𝑜𝑏𝑡𝑎𝑖𝑛 + 𝑠𝑖𝑔𝑛 𝑝ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒.
In the third step ;
1.
We generated a circuit in figure 18.6.(0.1 H , 0.1 mikroF and 1k ohm are parallel, 1 ohm are series) frequency become 1kHZ.
2.
We found phase difference between voltage and current
3.
10 notch => 360 degree 1 notch is 36 degree (current back , voltage is front )
4.
Voltage is 7.07 V rms , current is (0.02 V / √2 = 0.014 A rms
5.
Y x V = I , Y = (0.014 mA / 7.07 V ) = 2 mili Siemens
6.
Z = 1 / Y = 500 ohm

7.
Bc = 1/ Xc = 2*pi*f*C = 2*3.14*1KHz*(0.1 x 10 −6 ) = 0.62 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
8.
G = 1/R = 0.001 ohm
9.
Bl = 1/ Xl = 1/ 2*pi*f*L = 1 / 2*3.14*1KHz*0.1 = 1.59 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
10.
Bc-Bl = - 0.96 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
11.
Y = √𝐺 2 + (𝐵) 2
= 1.38 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠
12.
Z = 1/Y = 721 ohm
13.
B = - 0.96 x 10 −3 𝑠𝑖𝑒𝑚𝑒𝑛𝑠 and G = 1/R = 0.001 ohm
14.
𝜑 = arctan(− 0.96 x 10 −3 /0.001) = − 43 𝑑𝑒𝑔𝑟𝑒𝑒 (current is back , voltage is front)
15.
𝐼𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑠𝑜 𝑡ℎ𝑎𝑡 − 𝑠𝑖𝑔𝑛 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑠
COMMENTS
If only incutor used or capacitice used in the circuit or both of them used , what is changed in the circuit in terms of phase difference and voltage or current values and we leant which is front or back.We saw this changes.First experiment consists of series second circuit consists of parallel components.In addition ;
In the first part of the experiment , we firstly draw the circuit on CADET and measured some values that are phase difference voltage and current values.After that we calculated this values according to the formulas which we had learnt in before.Finally we compared this values.
Moreover, if inductor only used voltage is front , if capacitor is used voltage is back , if both are used according to the net reactance , we decide which is front or back.
In the second part of the experiment, we firstly draw the circuit on CADET and measured some values that are phase difference voltage and current values.We obtained some phase differences , if the circuit becomes inductive or capacitive.Our finding results are not totally the same , they are approximate equal each other.

Moreover, if inductor only used voltage is front , if capacitor is used voltage is back , if both are used according to the net reactance , we decide which is front or back.
We also draw the results in other three graph paper.
QUESTIONS
1.-- H ow did your calculated impedance magnitude in step 5 compare with impedance from the measured ac rms voltage and current ?
2k and 1.6k ohm we found
2 .-- We found 57.6 and 51.7 degrees , voltage is front of current.
3 .-- Our values are 796 ohm(X ) and 1278 ohm. Z = 2k ohm
4 .-- 27.6 degree and -38 degree are found and current is front according to the results.
5 .-- 1388 ohm and 2000 ohm we found
6 .- -Phase degrees are 36 degree (current is front ) and -43.94 (voltage is back ).We were expecting this results, because circuit is capacitive circuit voltage is back.
7 .- -First value is 500 ohm second one is 532 ohm
8 .- - Z = 500 ohm and calculated value = 622 ohm
9 .- -First phase difference is 60 degree and second one is we found 51 degree
10 .- - First result is 36 degree (current is front ) , second result is - 42 degree.We found current is back according to the oscilloscope screen , it is expected result.