Please do H I J R1CR2Figure 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 capacitore=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 tR1+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)+R22 (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 = + e4.(True, False)(j) The voltage across the switch is zero(True, False)

Answered: The switch is now closed[the switch has…

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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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)

Expert Solution

Part H

Let voltage across capacitor be V_C and that across resistor be V_R

Apply KVL in loop 2,

$V_{R} + V_{C} = 0 I R_{2} + \frac{q}{C} = 0 R_{2} \frac{d q}{d t} + \frac{q}{C} = 0 R_{2} \frac{d q}{d t} = - \frac{q}{C} \frac{d q}{q} = - \frac{d t}{R_{2} C} t a k i n g i n t e g r a l o n b o t h s i d e s; \int \frac{d q}{q} = - \int \frac{d t}{R_{2} C} \ln q = - \frac{d t}{R_{2} C} + C \ln q = - \frac{d t}{R_{2} C} + \ln K . . . (\sin c e K i s a c o n s \tan t, \ln K w i l l a l s o b e a c o n s \tan t) T a k i n g a n t i \log o n b o t h s i d e s; q = e^{- \frac{t}{R_{2} C} + \ln K} q = e^{- \frac{t}{R_{2} C}} e^{\ln K} q = K e^{- \frac{t}{R_{2} C}} r e p r e s e n t R_{2} C a s τ; q = K e^{- \frac{t}{τ}}$

Part H

$q = K e^{- \frac{t}{τ}} t a k i n g d e r i v a t i v e w i t h r e s p e c t t o t; I = \frac{d q}{d t} = - \frac{1}{τ} e^{- \frac{t}{τ}} = - \frac{1}{R_{2} C} e^{- \frac{t}{τ}}$

Voltage across the capacitor is;

$V_{c} = \frac{q}{C} = \frac{K e^{- \frac{t}{τ}}}{C} = K' e^{- \frac{t}{τ}} . . . . (\sin c e C i s a f i x e d v a l u e o f c a p a c i t o r, \frac{K}{C} w i l l a l s o b e a f i x e d v a l u e)$

Voltage across the resistance is;

$V_{R} = I R_{2} = - \frac{1}{R_{2} C} e^{- \frac{t}{τ}} \times R_{2} = - \frac{1}{C} e^{- \frac{t}{τ}} S i n c e t h e v a l i e o f c a p a c i t o r i s a f i x e d v a l u e, w e c a n r e w r i t e - \frac{1}{C} a s a c o n s \tan t v a l u e . = K " e^{- \frac{t}{τ}}$

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