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- he voltage drop across the capacitor rises from 0 to ℰ. Note that ℰ is never actually known in the measurement. In fact, the oscilloscope voltage is decalibrated, so that, whatever ℰ is, ℰ is at the top line while zero is at the bottom line. We don't measure voltage levels, but rather 1/2, 1/4, and 1/8 the maximum. Kirchhoff's voltage law give: ℰ = IR + Q/C or the following: dQdt=−1RC(Q−EC)dQdt=−1RC(Q−ℰC) The solution for the capacitor voltage is VC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC) The voltage is zero at t = 0, t is the rising time, and you have to know when the rising begins. Now, calculate VC (in dev) when t = RC ln 2. R = 0.6 kΩ C = 2.3 μFwouldn't it be -7.5J, if you consider ground as 0 and anythng higher is postive, the jar losses energy right?d. The Boolean equation y = ÃBC +Ã&ē + AB C TABC be implemented lesing only two - input NAND gates. The minimum number g gates requuêred is is to a) 3 (b, 4 (C, 5
- For multipole expansion, under what conditions, if any, is it an exact expression, and under what conditions, if any, is it an approximation? In what limits/cases is it the most useful?A dart hits the small square target shown below at a random point. Find the probability that the dart lands in the shaded circular region. Each side of the target is 15cm, and the radius of the shaded region is 5cm?