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- for the circuit shown below If v_in=5 sin2m10t then corner frequency will be ......Hz 1 UF 2 k Ohm 1 k Ohim 050 060 070 080 vinDiscussion 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 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
- 5. Suppose the following circuit is left open for a long time. 5000 22 24 V (a) What is the facial curren What is the carte II What is the ches What is the not the pos 80mF when we close thei 90s after we close the switch? dissipe the resistor 90s arter ene capacitor ons after we close the suit Close the switchi difference across the spacitor 008 after we close the switch across the resisto we close the swit 12 the potential (g) What is the energy stored in the capacitor 90s after we close the switch? (h) What is the energy dissipated by the resistor during the first 90s since the switch is closed?2_53528980697... -> For the circuit shown below, detemine the voltage of the -j10 2 capacitive reactance. -j5 0 ILOA(T) ==j102 jlon is Q 102 O 0.5L-90 52 A.What 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
- Draw a small signal AC equivalent circuit. Assume V=0 +20V Rc1 2k2 Cc 10µF R1 100k2 R Rc2 3k2 2k2 Vo 4µF Zo B = 100 Rs 1k2 R2 20k2 RE 5k2 CE 70µF VsFor the circuit shown in figure below ...select the value of сарacitance C2 Vcc +12 V Rc 1.0 kN C3 22 kN out Vin H) 2N3904 Vin = 100 mV f= 3 kHz R2 RE 8500 C2 6.8 kn 65.445 µF 61.545 µF 62.445 µF 62.554 µFZVO A VM ll l Ciassroo docs.google.com a Q1 For the following circuit diagram, 1 is .......... 50 15 V 12V 40 50 I2 70 60 90 12 a 4 A 1 O -1 احمد عصام احمد 8:00- äc lul 21.04.20 O D
- 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.Power factor 90 to 100% 88 to <90% 85 to <88% 80 to <85% 75 to < 80% 70 to < 75% 65 to < 70% 60 to <65% 55 to <60% 50 to <55% Less than 50% Surcharge None 2% 4% 9% 16% 24% 34% 44% 57% 72% 80% Use the table given for the following. A house with inductive loads has an avg consumption of 40kW and 26kVAR is supplied 230 Vrms source. We want to make the surcharge 0% by adding a capacitor. a. What is the smallest kVar of the capacitor that would reduce the surcharge to 0%?0 Consider the image below. Derive an expression for E as a function of the excitation voltage E¡ and the resistances R₁ and R₂. You may have to use Ohm's law and/or Kirchhoff's laws. Ei allt R1 Eo 0 R2