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- Fill in all the missing values. Refer to the formulas that follow. Resistance Capacitance Time constant Total time 150 k 100 F 350 k 35 s 350 pF 10 s 0.05 F 1.2 M 0.47 F 12F 0.05 s 86 k 1.5 s 120 k 470 pF 250 nF 100 ms 8 F 150 s 100 k 150 ms 33 k 4 F =RCR=CC=R Totaltime=5Find: Ib, Ic, Vce, Vb, Vc, Vrb Vre, & Vrc Note: show your solutions no solution means wrong answers. C2 HE 1000μF Rb 560 ΚΩ VCC 12V Rc 2kQ Q1 C1 요 HE 1000μF 2SC1815 R1 2000What is the total equivalent (total) inductance Lr of this circuit? 000 30mH L2 R 000 60mH 1 kQ 10 V L3 100mH 000 500mH 43 mH 720 mH O 173 mH o 日 点四RUA 36% DE 40°F PriSc Impr. Écr. F12 F10 F11 F7 F8 F9 F4 F5 F6 * & 3 0 1/4 1/2 4. 5 6. 7 8 2 R Y K L
- Discussion For the circuit below complete the table R 5 Ω 10 V 60 Hz 100 μF R Total 10+j0 10 0° Volts 68.623m +j364.06m Amps 370.5m Z 79.325° 5+ j0 0- j26.5258 5 - j26.5258 Ohms 520° 26.5258 -90° 26.993 Z-79.325°Discussion For the circuit below complete the table R 5Ω 10 V 60 Hz 100 μF R Total 10+j0 Volts 10 Z0° 68.623m +j364.06m Amps 370.5m Z 79.325° 5+ j0 0- j26.5258 5 - j26.5258 Ohms 520° 26.5258 -90° 26.993 Z-79.325°Discussion For the circuit below complete the table R 5 Ω 10 V 60 Hz 100 μF R Total 10+j0 10 0° Volts 68.623m+ j364.06m Amps 370.5m Z 79.325° 5+ j0 0- j26.5258 5 - j26.5258 Ohms 52 0° 26.5258 Z-90° 26.993 Z-79.325° N
- Question 13 For the circuit shown, the following mesh currents are given: IA =1.24271.2° A IB =0.45Z49.0° A Ic = 0.75Z150.2° A What is the value of the voltage VB? j5n + 20 VB 62- 45° V j4 N VA 240° V Vc 5230° V O 2.484- 108.8° V O 2.48471.2 V O 2.23/66.8° V O 2.23Z- 113.2° VFind equivalent capacitance for the following: a) C1 AHHHH C2 C3 C4 1µF 2µF 3uF 4µF b) C5 C9 1pF C6 2µF C10 3µF C7 4uF C11 5uF C8 6uF C12 A 7uF 8uF C29 с30 С35 C36 С37 С38 HHHH A С39 C32 42µF 42µF 1µF 2uF 3µF 4uF SuF 10μF C40 C31 C33 6pF 9uF C41 C34 ZuF 10μF C42 C43 8pF 8pF B. HHHHHDiscussion For the circuit below complete the table R 5Ω 10 V 60 Hz 100 µF R C Total 10 + j0 V Volts 10 Z0° 68.623m + j364.06m Amps 370.5m Z 79.325° 5+ j0 52 0° 5 - j26.5258 26.993 Z-79.325° 0 - j26.5258 Ohms 26.5258 Z-90°
- +5V VDO EN Xtal 8ה 7 +5V Freq1 HGND | GND х Mhz Sensor0 crystal Freq26 5 Vss Out 04 PRE Xtal 1k Output HA7210IP 4 CLK CLR SN74HC7491 GND 0.01 μ GND +5V 4 +5V +5V +5V OE O/P3 0. 1µF | 47µF GND SG351P |2 x MHz GND GND Consider the Quartz Crystal Microbalance Sensor shown above, with crystal resonant frequency, x = 3.3 Mhz. You are given the following values: Active crystal area = 1.6 cm², and the change in frequency = - 10 Hz Calculate the change in mass, Am in nano gram, up to 2 decimal places. Answer:9 62 802 5 mH 一 80V 156 1oov o=t エ has been n positien it moves to position The Switeh n the civauit in for a log time. At +=0 b. Find reltl for t70.نقطتان (2) E 6V * In the network shown in the figure, the magnitude of the current in (R1) using superposition theorem is R1 R2 8Q 8Q 3A 3A O 0.75A O 2A O 1.5 A O None of all O