14 V Μ Μ 6Ω Μ 5Ω Μ 3 Ω Μ Find the equivalent resistance seen by the source. [10.818 Ω ] ΤΩ
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- Inductive Circuits Fill in all the missing values. Refer to the following formulas: XL=2fLL=XL2ff=XL2L Inductance (H) Frequency (Hz) Inductive Reactance ( ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841Discussion 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 52 0° 0 - j26.5258 5 - j26.5258 Ohms 26.5258 Z-90° 26.993 L-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
- 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 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°ZVO 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
- Find: 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* Pzm is 220 2 V, Vz =20 V ZM=60 mA R21.2 kn V 60 W 12 W 12mw 1200 W What is the voltage value aer 0 0 0 0
- Find T 0.5 ms 40 92 www www 892 160 92 20 mH wwwDraw 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 VsThe discharging time constant for the circuit below is _nanoseconds. 5 V MPU 2 kQ RESET 100 kQ 1 pF SW 1