The series circuit has Vsupply 12Volts, R₁ =1K2, R₂ =2 KQ R3 = 3 KQ; • What is the power dissipater across all resistors? 24M 12mW a) 12 W 24mW Vsupply R₂ ww R₂

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Q: R₂=560K HE hie= 5002 hfe= 75 hoe=5μs Find: Zin, Zout, Av Vcc=50V RC= 3,5 k it out Rel=30052 Rez=5001

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Q: R1 Σ 10 ΚΩ R2 Μ HI Vcc = 10 v Rc 1 ΚΩ HI RE 560Ω

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Q: ww 20 Ω a o ww 10Ω ww- 80 Ω 60 2 50 Ω bo

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Q: How did they get a gain of 10?

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Q: what is the Q point

A: Given Data: R1 = 10k R2 = 2.2k Vcc = 25 V RE = 1k Rc = 3.6 k

Power Measurement

According to Blondel's theorem, the total power is measured with one less number of wattmeters than the number of wires present in the system. So, a single-phase and two-wire system will require one wattmeter and a single-phase and the three-wire system will require two wattmeters, and a three-phase and the three-wire system will also require two wattmeters and a three-phase and the four-wire system will require three wattmeter devices. The wattmeter is a device that measures power using one voltage and one current input as shown in the above figure. Multiple current or voltage input pair cables are used by many power analyzers and digital storage oscilloscopes (DSOs) for measuring watts, in effect within a single instrument acting as multiple wattmeters. It is possible to measure three-phase and four-wire power with one correctly specified power analyzer.