Experiment 5 Parallel RLC Circuits PDF

Experiment 5 Parallel RLC circuits Object: To study the characteristic of A.C. parallel circuits Apparatus: 1- Dual beam oscilloscope. 2- Function generator. 3- Resistance box. 4- Capacitance box. 5- Inductance box. 6- Coaxial cable. 7- Connecting wires. Theory A- Admittance: The admittance of a two...

Experiment 5 Parallel RLC circuits Object: To study the characteristic of A.C. parallel circuits Apparatus: 1- Dual beam oscilloscope. 2- Function generator. 3- Resistance box. 4- Capacitance box. 5- Inductance box. 6- Coaxial cable. 7- Connecting wires. Theory A- Admittance: The admittance of a two terminal network may be expressed as Y=I/V (siemen) Where Y=complex admittance V,I complex voltage, current The complex admittance is expressed as Y=G±JB Where :G=conductance B=suseptance B-Parallel RC Circuits For parallel RC circuit shown in Fig.(4). ⃑⃑ ⃑⃑⃑ ⃑⃑⃑ But there is 900 phase shift between resistance and capacitance currents , so we must first draw phasor diagram.In parallel circuits voltage can be consider as reference vector. phasor diagram can be drawing as shown in Figure (1) where source voltage VS is the reference vector (X-axis) and resistance current vector ⃑⃑⃑ is shown along the (X) axis , while the reactance current ⃑⃑⃑ is shown in the (Y) axis, since its current lead its voltage by 90°. Fig. 1: parallel RC circuit and its phasor diagram For currents For impedance | | | | ( ) | | | | ( ) | | √| | | | | | | | | √( | | ( ) | ) ( ) (| |) We must denoted that in parallel case we can't use oscilloscope in measuring phase angle. C- Parallel RL circuit A parallel RL circuit is illustrated in fig. 2. The admittance of the circuit is given by Y=YR+YL =I/R+1/jωL =G-jωL a) Circuit b) phasor diagram c) admittance triangle fig.2: parallel RL circuit Where, BL=1/(ɷL) and the angle Ф associated with Y is given by(see admittance triangle) Ф=tan-1[1/(ɷLG)] The current IR follow through the resistor is in phase with V and is given by IR = V/R= V.G (Amp) The current IL through inductor lags the voltage by 90° and is From the phasor diagram I=√ IL=V/XL=VBL = √( ) ( ) And the phase angle Ф is given by Ф=tan-1(-IL/IR) √ PROCEDURE Case (RC circuit ) 1. 2. 3. 4. Set the function generator to 8 VP.P(sine wave ,200 Hz) Use values R1=30Ω, and C=70µF. Connect the circuit shown in fig.3.1. Using voltmeter find IT, IR and IC 5. Measure the phase shift between applied voltage (v) and the current (i) by oscilloscope. 6. Find phase angle using one of the flowing ( ) ( ) ( ) 7. Draw the phasor diagram of the circuit 8. Compare the result obtained from step 6 with that obtained theoretically. Case (RL circuit ) 1. Repeat steps 1nd 2 in first case 2. Connect the circuit shown in fig.3.2. 3. Using voltmeter find IT, IR and IL 4. Measure the phase shift between applied voltage (v) and the current (i) by oscilloscope. 5. Find phase angle using a. ( ) ( ) ( ) 6. Compare the result obtained from step 5 with that obtained theoretically Discussion 1. Is the phase angle as that for RC series circuit.why? 2. 1- Is the phase angle as that for RL series circuit.why? 3. Is the conductance of the coil G the same as 1/r where r is the 4. resistance of the coil ? which is easier to measure experimentally? 5. Comment on the result obtained.