Charging Capacitor: For a charging capacitor the Kirchoff's Loop Rule gives R E - IR- = 0 C C In this case the current is entering the positive plate so I = dQ/dt = CdAVc/dt and we get dVe V. = 0 dt E – RC 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 2 20 40 60 80 100 time (ms) b. What is the current at t = 10 ms? Hint: the easiest way to do this is to use the loop rule. V. (volts)

Charging Capacitor: For a charging capacitor the Kirchoff's Loop Rule gives R E - IR- = 0 C C In this case the current is entering the positive plate so I = dQ/dt = CdAVc/dt and we get dVe V. = 0 dt E – RC 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 2 20 40 60 80 100 time (ms) b. What is the current at t = 10 ms? Hint: the easiest way to do this is to use the loop rule. V. (volts)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

100%

Charging 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.

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