Question

Chapter 26 Laboratory Application Assignment In this lab application assignment you will examine an RC coupling circuit and an RC low-pass filter. In the RC coulina circuit you will see how the series capacitor blocks the de component of the input voltage but passes the ac component. In the RC low-pass filter you will see how the low frequencies are passed from input to output with little or no attenuation but the higher frequencies are severely attenuated or blocked. Equipment: Obtain the following items from your instructor • Function generator • Oscilloscope • DMM • 0.1- F and 0.22-uF capacitors • 2.2-k2 and 10- carbon-film resistors What value of rms voltage would you expect to measure across inout terminals 1 and 2 V -LEX How much de voltage would you expect to measure across the capacitor, C? V S V How much dc voltage would you expect to measure across the resistor, R? V O V dc. Construct the circuit in Fig. 26-430. With a DMM connected to the output of the function generator, adjust the de offset control to obtain a de value of +5 V dc. Also, while viewing the oscilloscope, adjust the amplitude and frequency controls of the function generator to obtain an output voltage of 10 V, with a frequency of 1 kHz. Have your instructor check your settings Next, measure and record the following values in Fig. 26-430: V .. VO ... Vald - How do these values compare to those predicted2 RC Coupling Circuit Examine the RC coupling circuit in Fig. 26-430. Notice the input voltage is a pulsating de voltage whose value remains entirely nositive. The input waveform (across terminals 1 and 2) is shown in Fig. 26-43b. The output from the RC coupling circuit is taken across terminals 3 and 4, which is across the resistor, R. What value of de voltage would you expect to measure across input terminals 1 and 2? Vind _SV 0 in Fig. 26-430, calculate X. and Zat 1 hey Z Next, calculate and record the following rms values: - ..- 8.665 .. . Using your DMM, measure and record the following rms values: V = .V

checking to see if the answers i got are correct and help with the other parts. thank you

Answer 1

# RC coupling Circuit CE 0-22-4F Vout 2p.p=101 3 RAJOK- Vde=svų 12 j WC voltage doross terminal land 2 Between 1 and 2, One dc Source and one ac source is present: • BIC (average) value of AC source = 0 volt I because positive part equals to negative part during one cycle} o occaverage) value of dc source - 5 volt Total de across 1 and 2 = 5+0 Vacci-2) = 5 volt iRMS Voltage dcross terminal 1 and 2 1 maximum value of Ac source = 10 - 5 volt RMS Value of Ac source = Vm= = 3.54 Volt Vormsi = 3-54 volt] → RMS Value of dc source = average voltage Vams2 = 5 Volt Total RMS = s výmsi + vaims2 ()2+(5)2 = 1375 = 6.123 Verms (1-2) = 6:123 volt]

hi) uc Voltage dorass Capacitor- Capacitore charging and discharging will happen when connected with AC sowice. So average dc will be zeno. o Due to DC offset, capacitou will get charged up to dc value Vdc across capaciter - 5 volt [Vdc = 5 volt ) co citoy iv) UC voltage across Resistore when capacitore fully charged upto maximum de valué, it will act as open-circuit and de voltage ACHOSS resistor will be zero. Vdc=07 doross resistor uy calculate x and z Kc = 20$C - 2771000x 0 22x1076 -1 1-3823x10-3--0723.43-2 x = -17234343 2=R-jxc (Rand (in series) 2: 10000 -j 723.43 = 10026.134-4-137° Fz= 10026-132-4137°