Given the plot of Ve (t), i.e., the voltage on a capacitor, which is the correct expression forIc (t), i.e., the current on the capacitor.SV0Ve(t)t = 10s t=20s t=30sOHint: Ic (t) =dv. (t)dt-, i.e., Ic (t) =-For instance, from t=0 to t=10s, IcIc(t)=0.5A,OA,-0.5A,0.5A,1c(t) = 0.55A,-0.5A,0.5A,t=40sAELt € [0,10)t € [10,30)t € [30,40)te [0,10)t € [10,30)te [30,40)te [0.10)le(t)=0.25A, te [10,30)-1230 103AV.(t)AttV. (10) V. (0)10-0[mm]5-010-00.54 Hint: Ic (t) = 2, i.e., Ic (t) =dVc(t)dtFor instance, from t=0 to t=10s, IcOIc(t) =0.5A,OA,-0.5A,0.5A,Ic(t) = 0.55A,-0.5A,0.5A,Ic(t) = 0.25A,0.5A,0.5A,1c(t) = 0.55A,1A,t € [0,10)t € [10,30)t € [30,40)te [0,10)t € [10,30)t € [30,40)t € [0,10)t € [10,30)t € [30,40)t = [0,10)te [10,30)AV.(t)Att € [30,40)=Vc(10)-Vc (0)10-0=5-010-0: 0.5A=

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Given the plot of V. (t), i.e., the voltage on a capacitor, which is the correct expression for Ic (t), i.e., the current on the capacitor. SV 0 1 Vc(t) t=10s t=20s t = 30s t = 40s Hint: Ic (t) = , i.e., Ic (t) = dVc (t) dt O For instance, from t=0 to t=10s, Ic = Ic(t)= 0.5A, OA, -0.5A, 0.5A, Ie(t)= 0.55A, -0.5A, t = [0,10) t € [10,30) AV.(t) At t € [30,40) te [0,10) te [10,30) te [30,40) t Vc (10)-V.(0) 10-0 Came 5-0 10-0 - 0.5A

Given the plot of V. (t), i.e., the voltage on a capacitor, which is the correct expression for Ic (t), i.e., the current on the capacitor. SV 0 1 Vc(t) t=10s t=20s t = 30s t = 40s Hint: Ic (t) = , i.e., Ic (t) = dVc (t) dt O For instance, from t=0 to t=10s, Ic = Ic(t)= 0.5A, OA, -0.5A, 0.5A, Ie(t)= 0.55A, -0.5A, t = [0,10) t € [10,30) AV.(t) At t € [30,40) te [0,10) te [10,30) te [30,40) t Vc (10)-V.(0) 10-0 Came 5-0 10-0 - 0.5A

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

Given the plot of Ve (t), i.e., the voltage on a capacitor, which is the correct expression for
Ic (t), i.e., the current on the capacitor.
SV
0
Ve(t)
t = 10s t=20s t=30s
O
Hint: Ic (t) =
dv. (t)
dt
-, i.e., Ic (t) =
-
For instance, from t=0 to t=10s, Ic
Ic(t)=
0.5A,
OA,
-0.5A,
0.5A,
1c(t) = 0.55A,
-0.5A,
0.5A,
t=40s
AEL
t € [0,10)
t € [10,30)
t € [30,40)
te [0,10)
t € [10,30)
te [30,40)
te [0.10)
le(t)=0.25A, te [10,30)
-1230 103
AV.(t)
At
t
V. (10) V. (0)
10-0
[mm]
5-0
10-0
0.54

Hint: Ic (t) = 2, i.e., Ic (t) =
dVc(t)
dt
For instance, from t=0 to t=10s, Ic
O
Ic(t) =
0.5A,
OA,
-0.5A,
0.5A,
Ic(t) = 0.55A,
-0.5A,
0.5A,
Ic(t) = 0.25A,
0.5A,
0.5A,
1c(t) = 0.55A,
1A,
t € [0,10)
t € [10,30)
t € [30,40)
te [0,10)
t € [10,30)
t € [30,40)
t € [0,10)
t € [10,30)
t € [30,40)
t = [0,10)
te [10,30)
AV.(t)
At
t € [30,40)
=
Vc(10)-Vc (0)
10-0
=
5-0
10-0
: 0.5A
=

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