)In a voltage divider network of FET, if the resistors R1 and R2 are equal VG = VCC VG = ½ Vcc VG = 0V VG = VGS
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1)In a voltage divider network of FET, if the resistors R1 and R2 are equal
VG = VCC
VG = ½ Vcc
VG = 0V
VG = VGS
Consider the formula for the gate voltage in the FET devices is,
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- Consider four resistors in parallel connection: R1 = 3520.45 Q, R2 = 33.09 Q, R3 = 1484.22 Q and R4 = 11.83 Q. What is the equivalent resistance? R2 R3 R RT O 11.24 O 7.78 19.88 O 8.64 O 6.91OCX elp VCC VEE 6V 3V R2 A R1 2kQ 2kQ R5 2kQ a) Find the voltage at point A due to VCC only. b) Find the voltage at point A due to VEE only. c) Find the voltage at point A when both are activated.Employing the characteristic curve in Figure 1 and obtain the design for a voltage divider configuration that has a Q point of ICQ = 5 mA and VCEQ = 8 V. Using VCC = 24 V and RC = 3RE. Find the following: a) Draw the configuration indicating each of the elements b) Determine RC and RE c) Find VE d) Determine VB e) Calculate ? for point Q f) Find R2 if R1 = 24 kΩ
- For R1=9, R2=1, R3=1, R4=1 and 15-13.5 A in the shown circuit, use current divider to find the following: 15 i1 (in Ampere) = a. 10.125 b. 0.48214285714286 c. 10.125 d. 3.375 i2 (in Ampere) = a. 5.6410714285714 b. 4.3392857142857 c. 1.125 d. 2.475 i3 (in ampere) = RI i4 (in ampere) = R2< 1₂ R32 13 R4 isO For the voltage -divider bis confgurahon in Figure I, determine : o16V a) IBe b) Ice c) VCeQ 62ka. 33.9kn LICQ o Ve Figure 1 d) Ve VB VCEQ B = 80 e) Ve f) VB IBQ oVE 0.68 KAThe voltage divider shown in the figure has a potential difference (V) between point A and B and a potential difference (VO) between points C and D. What is V FO? A B RM 2R O A. 2 OB. 4 O C.3 OD. S M 2R R :2R C D
- 1. For the voltage-divider bias configuration, determine: a. IBQ: b. Icq. c. VCEQ- d. Vc. e. VE - f. VB 16 V 3.9 k2 ICQ 62 kN Vc + VB VCEQ B = 80 IBQ VE 9.1 k2 0.68 k2One of the resistors in this voltage divider circuit has failed (either open or shorted). Based on the voltage readings shown at each load, comparing what each load voltage is versus what it should be, determine which resistor has failed and what type of failure it is: - 50 V R₁ wwwww 30 R₁ w... R4 Load #1 Design voltage: -9 V Actual voltage: -31.7 V Load #2 Design voltage: -18.7 V Actual voltage: -33.8 V Load #3 Design voltage: -35.3 V Actual voltage: -41.1 VQuiz nav Question 6 R2 Not yet answered R1 Marked out of ww Finish atte- 3.00 R3 RA F Flag question Rs Rs In the above circuit, R1 = 100, R2 = 200, RB= 150 R4 = 5O R5 = 300 ad R6 = 400. The battery voltage is 9V 1 The equivalent resistande of the dircuit is 2. The source current is 3. The voltage drop across RS is
- The voltage divider rule states that Select one: a. the voltage applied to a parallel resistīve circuit will be dropped across all resistors in proportion to the magnitude of the individual resistors b. the voltage applied to a serles resistive circuit will be dropped across all resistors in proportion to the magnitude of the individual resistors c. the current passing to a series resistive circuit will be dropped across all resistors in proportion to the magnitude of the Individual resistors d. None of the answersFor the voltage divider bias JFET circuit shown, What is the drain current Ip if VD = %3D 8V? VDD = 16V R1 8M2 Ro 1k2 R2 1M2 Rs 2002 a. 0.63mA b. 8mA O C. 6.3mA d. 15.2mA%3D ) Drurde (l0o1102 and (o