The switch is now closed[the switch has no resistance] (h) R2 t + & = 0 q = Ce • Vc = ce¯# ; VRa = IR2 = ce;r'= R2C (True, False) .13= %3! (i) The current through the switch is I = I1 + 12 = + e7 (True, False) (i) The voltage across the switch is zero (True, False)
The switch is now closed[the switch has no resistance] (h) R2 t + & = 0 q = Ce • Vc = ce¯# ; VRa = IR2 = ce;r'= R2C (True, False) .13= %3! (i) The current through the switch is I = I1 + 12 = + e7 (True, False) (i) The voltage across the switch is zero (True, False)
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter21: Resistive-capacitive Series Circuits
Section: Chapter Questions
Problem 6RQ: A 15-F AC capacitor is connected in series with a 50 resistor. The capacitor has a voltage rating...
Related questions
Question
Please do H I J
![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%2F151adc3a-526b-4ffa-85c0-d1ab0551687d%2Fsreaj1e_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)
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Part H
Let voltage across capacitor be VC and that across resistor be VR
Apply KVL in loop 2,
Part H
Voltage across the capacitor is;
Voltage across the resistance is;
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