Experiment No. 3: RLC Series Circuits PDF

Experiment no.3 R L C series circuits Object: To study the characteristics of ac 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- Impedance: the impedance of a two termin...

Experiment no.3 R L C series circuits Object: To study the characteristics of ac 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- Impedance: the impedance of a two terminal network may be expressed as: Z=v/I (ohms) Where z = complex impedance V= complex voltage I complex current The complex impedance is also expressed as: Z=r+jx in Cartesian form Z=zeiФ in polar form. The value of Ф can measure based on the value of the T and t , where, T is the period time of the wave, and t is the time shift between the signal on the resistance or the inductor as shown in the figure below. Phase angle in degree: scale. Where T, t is distance in any length Phase measuring using two channels of oscilloscope B- THE SERIES RC CIRCUIT: Fig 3.1 and fig 3.2 illustrated a series rc circuit connected to an ac voltage source. The applied voltage can be expressed as: V=Vm sinωt =Ir + ∫ 𝑖𝑑𝑡 The solution of this differential equation results in. I=IM sin(ωt+Ф) This equation shows that current I leads voltage v by Ф The impressed voltage v can expressed form as: V=VR - jVC Which can be expressed as (see phase diagram): |v|=√ √ =|I| √ Where XC=1/(ɷC) And the phase angle Ф is given by: Ф=tan-1(-VC/VR)=tan-1(-XC/R) Fig 3.1 : Circuit of series RC and wave form of the circuit Fig. 3.2 phasor diagram of series RC C- Series RL circuit: Fig.3.3 illustrate a series RL circuit to which a sinusoidal voltage is impressed. v=VM sin ωt Fig 3.3: circuit and waveforms Applying K.V.L, one gets: V=IR- Ldi/dt Solving that equation for I, one gets I=IM Sin(ɷt-Ф) Thus the current I lags the voltage v by Ф as shown in fig 3.4. From phase diagram, one express: |v|=√ √ =|I| √ Where XL= ωC Ф=tan-1(VL/VR)=tan-1(XL/R) Fig 3.4: Phasor diagram of a series RL circuit Procedure Part A: series RC circuit 1. Connect the circuit shown in fig 3.1 with (R= 100 Ω , 10 µ F) 2. Set the input voltage at 3VP.P, 300Hz 3. Measure the phase shift between the current I and the applied voltage v by using oscilloscope. Vp.p Freq. Draw the input signal Draw output signal on (R&C) 4. Repeat all the previous steps with 3 Vp.p , 1KHz. Vp.p Freq. Draw the input signal Draw output signal on (R&C) 5. Calculate the phase angle theoretically for the two cases Part B:series RL Circuit 1. Connect the circuit shown in fig.3.3 with (R=100 Ω, 10 mH) 2. Set input voltage at 4VP.P, 250Hz 3. Measure the phase shift between the current I and the applied voltage v by using oscilloscope. Vp.p Freq. Draw the input signal Draw output signal on (R&C) 4. Repeat all the previous steps with 4 Vp.p , 1KHz. Vp.p Freq. Draw the input signal Draw output signal on (R&C) 5. Calculate the phase angle theoretically for the two cases Discussion 1- Explain why the phasor and the impedance have the same angle. 2- Compare the result for utilized different frequency in same circuit for part A and B 3- At what condition the following results obtained? a- Phase angle equal zero b- The applied voltage lead the current by 90°. c- The average power equal to zero. 4- In general, how would the phasor diagram of Figure 3.2 change if the frequency was raised? 5- In general, how would the phasor diagram of Figure 3.4 change if the frequency was lowered?