What is Kirchhoff's 1st equation for this junction?
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A: Given: R =0.7 KΩ C = 6.3 μF
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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 μFWhat is the difference between conductors and insulators from the atomic perspective?Describe the processes that occur at a metallic electrode surface in response to a step- change in the applied potential. Your answer should include non-Faradaic and Faradaic processes.