Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- Please assist with this practice problem 1 c,d,e with details on how to do it. Thank you.arrow_forward6) In the circuit below, the constant-amplitude voltage generator delivers four times more current at very low frequencies than it does at very high frequencies. If R1=100 N, find the resistance R2. (HINT: Consider how an inductor and a capacitor behave at high and low frequencies.) R1 V C R2arrow_forwardCharging Capacitor: For a charging capacitor the Kirchoff's Loop Rule gives R E – IR C ww C In this case the current is entering the positive plate so I = dQ/dt = CdAVc/dt and we get dVc E – RC - Vc = 0 dt The solution to the differential equation is V.(t) = E (1 – e-t/(RC)) (charging capacitor) Notice that V.(0) = 0 and V.(0) = E as we expect for a charging capacitor. 5. You can easily find the time constant if you are given a graph of voltage across a charging capacitor as a function of time. Whent = RC, the voltage across the capacitor is V.(t = RC) = E(1 – e-1) × 0.63 E. Therefore the time constant is just how long it takes for AV(t) to reach 63% of the EMF. The graph shows the voltage across a charging capacitor as a function of time. The resistance of the circuit is 7.5 kN. а. Determine the capacitance of the capacitor. 10 8 60 40 time (ms) 20 80 100 b. What is the current at t = 10 ms? Hint: the easiest way to do this is to use the loop rule. (volts) 4.arrow_forward
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