Sive the symbolic expression for the emf Eusing KVL for the circuit with S, closed and S2 open. Give your answer in terms of the current I, resistor R capacitors C1 and C2 and charges stored in the respective capacitors Q1 and Q2. Jse to denote product and / to denote division. So to group the product of, say, a and b_1 write a*b_1. And to write a ratio of say, c_1 and d write _1/d. To add the product and ratio write a*b_1 + c_1/d. aWrite the mathematical expression for emf E. n the figure there's a circuit with an emf E = 15 V, two resistors R1 witches S1 and S2. = 11 kNand R2 = 3.3 kN two capacitors C = 29 µFand C2 = 29 µFand two ) Find the time constant for this configuration of the circuit.

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Solve d, e, and f. 

PartI
a
S.
3
Give the symbolic expression for the emf Eusing KVL for the circuit with S1 closed and S2 open. Give your answer in terms of the current I, resistor R,
capacitors C1 and C2 and charges stored in the respective capacitors Q1 and Q2.
Use * to denote product and / to denote division. So to group the product of, say, a and b_1 write a*b_1. And to write a ratio of say, c_1 and d write
c_1/d. To add the product and ratio write a*b_1 + c_1/d .
aWrite the mathematical expression for emf E.
E =
= 29 µFand two
In the figure there's a circuit with an emf E = 15 V, two resistors R1 = 11 kNand R2 = 3.3 kN two capacitors C1
switches S, and S2.
29 μFand C,
=
b) Find the time constant for this configuration of the circuit.
Time constant T
Give your answer up to at least three significance digits.
ms
c) Find how much charge will be stored in C, after time t = 1.4Tseconds.
Charge stored in C2
Give your answer up to at least three significance digits.
µC
After t = 10Tseconds, we open switch S1 and close switch S2. Mark current time as t' = 0 In this configuration, capacitor C2 discharges through the
resistor R2.
d) Find the charge stored on the capacitor C, as the switch S, was turned off.
Charge stored
Give your answer up to at least three significance digits.
e) Find the current throught the resistor R2 at time t' = 1.5T
Current
Give your answer up to at least three significance digits.
mA
f) Find the time from t' =
9 that will be taken by the capactor to loose half of its charge.
Time taken to discharge
Give your answer up to at least three significance digits.
ms
Transcribed Image Text:PartI a S. 3 Give the symbolic expression for the emf Eusing KVL for the circuit with S1 closed and S2 open. Give your answer in terms of the current I, resistor R, capacitors C1 and C2 and charges stored in the respective capacitors Q1 and Q2. Use * to denote product and / to denote division. So to group the product of, say, a and b_1 write a*b_1. And to write a ratio of say, c_1 and d write c_1/d. To add the product and ratio write a*b_1 + c_1/d . aWrite the mathematical expression for emf E. E = = 29 µFand two In the figure there's a circuit with an emf E = 15 V, two resistors R1 = 11 kNand R2 = 3.3 kN two capacitors C1 switches S, and S2. 29 μFand C, = b) Find the time constant for this configuration of the circuit. Time constant T Give your answer up to at least three significance digits. ms c) Find how much charge will be stored in C, after time t = 1.4Tseconds. Charge stored in C2 Give your answer up to at least three significance digits. µC After t = 10Tseconds, we open switch S1 and close switch S2. Mark current time as t' = 0 In this configuration, capacitor C2 discharges through the resistor R2. d) Find the charge stored on the capacitor C, as the switch S, was turned off. Charge stored Give your answer up to at least three significance digits. e) Find the current throught the resistor R2 at time t' = 1.5T Current Give your answer up to at least three significance digits. mA f) Find the time from t' = 9 that will be taken by the capactor to loose half of its charge. Time taken to discharge Give your answer up to at least three significance digits. ms
Expert Answer
Step 1 Analyzing the question for part I
"Since you have asked multiple question, we will solve the first question for you. If you want any specific
question to be solved then please specify the question number or post only that question."
So, solving part I we have:
(a) Switch S, is closed and S2 is open. The circuit is thus a series RC circuit in which the two capacitors are in
series whose equivalent is:
Ceg = C1 * C2/(C¡ + C2)
(1)
The
e E will divide itself between the two capacitances and so in steady state current will be zero in the
circuit. Thus we have expression of E as:
E(1) = E(1-e ) volts
(2)
where E(t) is total voltage across both the capacitors and Ceg can be substituted from (1)
Step 2 Answer of the question (b) and (c)
(b) Since switch Sj is closed and S2 is open the time constant for the circuit is given as:
T = CeqR
Equivalent capacitance is given as:
Ceg=(29 * 29)/(29 + 29)
14.5μF
The time constant is given as:
T=14. 5 × 10-6 × 11 × 10³
=0. 1595 sec
=159. 5 ms
(c) The total voltage of both the capacitor will be equal to:
E(t) = 15(1 – e-) volts
Putting value of time we get:
1.4r
E(t)=15(1 – e¯
=11.301V
The voltage across C2 can be calculated by voltage division rule as:
V;=E(t) ×
C+C2
29
=11. 301 x
29+29
=5. 6505V
The charge stored in C2 will be given as:
Q2=C2V2
=29µF x 5. 6505
=163. 865 µC
Thus all the answers of part I have been calculated.
Transcribed Image Text:Expert Answer Step 1 Analyzing the question for part I "Since you have asked multiple question, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question." So, solving part I we have: (a) Switch S, is closed and S2 is open. The circuit is thus a series RC circuit in which the two capacitors are in series whose equivalent is: Ceg = C1 * C2/(C¡ + C2) (1) The e E will divide itself between the two capacitances and so in steady state current will be zero in the circuit. Thus we have expression of E as: E(1) = E(1-e ) volts (2) where E(t) is total voltage across both the capacitors and Ceg can be substituted from (1) Step 2 Answer of the question (b) and (c) (b) Since switch Sj is closed and S2 is open the time constant for the circuit is given as: T = CeqR Equivalent capacitance is given as: Ceg=(29 * 29)/(29 + 29) 14.5μF The time constant is given as: T=14. 5 × 10-6 × 11 × 10³ =0. 1595 sec =159. 5 ms (c) The total voltage of both the capacitor will be equal to: E(t) = 15(1 – e-) volts Putting value of time we get: 1.4r E(t)=15(1 – e¯ =11.301V The voltage across C2 can be calculated by voltage division rule as: V;=E(t) × C+C2 29 =11. 301 x 29+29 =5. 6505V The charge stored in C2 will be given as: Q2=C2V2 =29µF x 5. 6505 =163. 865 µC Thus all the answers of part I have been calculated.
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