In the RLC circuit shown below, determine VR1 and VL1 when VC1 =4e-3000(t) Hint: See file "EET111 Solved examples" in your Canvas class under "Additions Resources" di1 Find i1= C1 and VR1 = R1 x il dt R2 v(t) dVc1 dt C2 HE 1μF Then, VL1: = R1 ww 1.5ΚΩ L2 1mH L1 1kQ L1 15mH i1 C1 0.5μF
In the RLC circuit shown below, determine VR1 and VL1 when VC1 =4e-3000(t) Hint: See file "EET111 Solved examples" in your Canvas class under "Additions Resources" di1 Find i1= C1 and VR1 = R1 x il dt R2 v(t) dVc1 dt C2 HE 1μF Then, VL1: = R1 ww 1.5ΚΩ L2 1mH L1 1kQ L1 15mH i1 C1 0.5μF
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
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I got stuck on the second part, and didn't make it to the third part. Thanks!
![In the RLC circuit shown below, determine VR1 and VL1 when VC1 =4e-3000(t)
Hint: See file "EET111 Solved examples" in your Canvas class under "Additions Resources"
di1
Find i1= C1
and VR1 = R1 x il
dt
R2
v(t)
dVc1
dt
C2
HE
1μF
Then, VL1: =
R1
1.5ΚΩ
L2
1mH
L1
1kQ
L1
mon
15mH
i1
C1
0.5μF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F670d8787-56ca-4341-b97d-e84fef52a8ee%2F56516611-a34e-46eb-aea5-5ff984b2a2fe%2Fazr1ui_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In the RLC circuit shown below, determine VR1 and VL1 when VC1 =4e-3000(t)
Hint: See file "EET111 Solved examples" in your Canvas class under "Additions Resources"
di1
Find i1= C1
and VR1 = R1 x il
dt
R2
v(t)
dVc1
dt
C2
HE
1μF
Then, VL1: =
R1
1.5ΚΩ
L2
1mH
L1
1kQ
L1
mon
15mH
i1
C1
0.5μF
![= C₁
i₁ =
dv₁
dt
VL1 = L₁
|
di
dt
=
d
dt
(0.5µF) (4e¯3000(t)v) = (0.5µF)(4V)(–3000)e¯³
-3000(t) = -6000e-3000 (t) μA
d
dt
(15mH) -6000e-3000(t) μA) = (15mH) (-6000µA) (-6000Hz)(e-3000 (t))](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F670d8787-56ca-4341-b97d-e84fef52a8ee%2F56516611-a34e-46eb-aea5-5ff984b2a2fe%2F97dorna_processed.jpeg&w=3840&q=75)
Transcribed Image Text:= C₁
i₁ =
dv₁
dt
VL1 = L₁
|
di
dt
=
d
dt
(0.5µF) (4e¯3000(t)v) = (0.5µF)(4V)(–3000)e¯³
-3000(t) = -6000e-3000 (t) μA
d
dt
(15mH) -6000e-3000(t) μA) = (15mH) (-6000µA) (-6000Hz)(e-3000 (t))
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