3. In figure 3 before the switch is closed: (a) (R1 + R2)a + = - q = Ce(1 - e ) - 1 = - Vc = c(1 - e) ; VR, = IR, = (R1 + R2)C (True, False) (b) The displacement current Ia through the capacitor is equal to the conduction current through the wires i.e. Ia = RtR (True, False) (c) The time it takes for the voltage across capacitor to reach half of its maximum value is t 1 = (R1 + R2)C lIn 2 (True, False)
3. In figure 3 before the switch is closed: (a) (R1 + R2)a + = - q = Ce(1 - e ) - 1 = - Vc = c(1 - e) ; VR, = IR, = (R1 + R2)C (True, False) (b) The displacement current Ia through the capacitor is equal to the conduction current through the wires i.e. Ia = RtR (True, False) (c) The time it takes for the voltage across capacitor to reach half of its maximum value is t 1 = (R1 + R2)C lIn 2 (True, False)
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
Please do 3 A, B, C, D
![R1
C
R2
Figure 3:
3. In figure 3 before the switch is closed:
(a) (R1 + R2) 44 + 4 = e - q = Ce(1 - eT) - I = Ri Roe
- Va = e(1 - e7) ; VR,
IR; =
;T =
(R1 + R2)C (True, False)
(b) The displacement current Ia through the capacitor is equal to the conduction current through the wires i.e. Ia = Raikae
(True, False)
(c) The time it takes for the voltage across capacitor to reach half of its maximum value is t1 = (R1 + R2)C In 2 (True, False)
(d) The power delivered by the battery is Pe = Ie = RRaei and energy delivered by the battery at time t is E = S Pedt = eq =
?C(1 - e) (True, False)
(e) Energy Stored in the capacitor at time t is Ec =
cv? and the power associated with the capacitor
e
=D 융 0 = Veld =D €(1-e"
= CVc YE (True, False)
(f) The power consumed by the two resistors is PR = 12 (R1 + R2) =
2* and the energy dissipated by the two resistors at time t
R1+R2 €
is ER1+R2 = Só Pdt =
(g) If (e) is true the total energy dissipated by the resistors is independent of the their resistance and is equal to Et-)
- e
-#) (True, False)
+R2
2 (True, False)
The switch is now closed[the switch has no resistance]
(h) R2 d4 + = 0 - q = Cee T -I = Te
- Vc = ce
; VR2 = IR2 = ce
*; r' = R2C (True, False)
(i) The current through the switch is I = I + 12 = + e
4.
(True, False)
(j) The voltage across the switch is zero
(True, False)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90ba196a-a4da-4c5b-97fd-f3d58be17337%2F33f15cbb-c066-40c2-8ead-f359d7f0ef25%2Fjqxl59b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:R1
C
R2
Figure 3:
3. In figure 3 before the switch is closed:
(a) (R1 + R2) 44 + 4 = e - q = Ce(1 - eT) - I = Ri Roe
- Va = e(1 - e7) ; VR,
IR; =
;T =
(R1 + R2)C (True, False)
(b) The displacement current Ia through the capacitor is equal to the conduction current through the wires i.e. Ia = Raikae
(True, False)
(c) The time it takes for the voltage across capacitor to reach half of its maximum value is t1 = (R1 + R2)C In 2 (True, False)
(d) The power delivered by the battery is Pe = Ie = RRaei and energy delivered by the battery at time t is E = S Pedt = eq =
?C(1 - e) (True, False)
(e) Energy Stored in the capacitor at time t is Ec =
cv? and the power associated with the capacitor
e
=D 융 0 = Veld =D €(1-e"
= CVc YE (True, False)
(f) The power consumed by the two resistors is PR = 12 (R1 + R2) =
2* and the energy dissipated by the two resistors at time t
R1+R2 €
is ER1+R2 = Só Pdt =
(g) If (e) is true the total energy dissipated by the resistors is independent of the their resistance and is equal to Et-)
- e
-#) (True, False)
+R2
2 (True, False)
The switch is now closed[the switch has no resistance]
(h) R2 d4 + = 0 - q = Cee T -I = Te
- Vc = ce
; VR2 = IR2 = ce
*; r' = R2C (True, False)
(i) The current through the switch is I = I + 12 = + e
4.
(True, False)
(j) The voltage across the switch is zero
(True, False)
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