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Q4. For the circuit below, determine the following: a. The time constant of the circuit when the switch is in position 1. b. Find the time constant when the switch is in position 3. c. Find the voltage vc after the switch has been in position 1 for 60 ms. d. After the switch has been in position 1 for 60 ms, it is moved to position 2. What is the value of vc after 600 ms? Vo 20 3 10 µF 2 ko 100 V 4 k2 3 k.

Q4. For the circuit below, determine the following: a. The time constant of the circuit when the switch is in position 1. b. Find the time constant when the switch is in position 3. c. Find the voltage vc after the switch has been in position 1 for 60 ms. d. After the switch has been in position 1 for 60 ms, it is moved to position 2. What is the value of vc after 600 ms? Vo 20 3 10 µF 2 ko 100 V 4 k2 3 k.

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Please answer D only, i already have answer for a, b, and c, thanks.

i am attaching answer a, b and c for your reference.

Q4. For the circuit below, determine the following: The time constant of the circuit when the switch is in position 1. b. Find the time constant when the switch is in position 3. c. Find the voltage vc after the switch has been in position 1 for 60 ms. d. After the switch has been in position 1 for 60 ms, it is moved to position 2. What is the value of vc after 600 ms? a. 1 20 10 µF 2 ka 3 100 V 4 k2 3 ko
Transcribed Image Text:Q4. For the circuit below, determine the following: The time constant of the circuit when the switch is in position 1. b. Find the time constant when the switch is in position 3. c. Find the voltage vc after the switch has been in position 1 for 60 ms. d. After the switch has been in position 1 for 60 ms, it is moved to position 2. What is the value of vc after 600 ms? a. 1 20 10 µF 2 ka 3 100 V 4 k2 3 ko
More than three subparts were required to be answered, so the 1st three has been answered. From the working principle of a capacitor it is understood that the capacitor offers zero resistance when it is uncharged and in that condition it can be taken as short circuit. When the capacitor is fully charged it will offer infinite resistance. When the position of the switch is at position 1; the capacitor will be charging from a 100 V battery; Applying Kirchhoff's Voltage law - - i (R, + R2) + V = 0 R = 2 kilo ohm, R2 = 3 kilo ohm and R3 = 4 kilo ohm R = R + R2 = 2+3 = 5 kilo ohm = 5 x 10 ohm V = 100 V The capacitor C = 10 µF -- iR + V = o i = 4 -온-R + V=0 V - = R di Now integrating the above equation, we get () q = CV (1 - e CR) CR = T = time cons tan t = 10 x 10-6 x 5x 10 = 0.05 s Answer : The cons tan t of the circuit when it is at position 1 is 0. 05 s Step 2 When the switch is at position 3, we again apply the Kirchhoff's Voltage law to the circuit, This is the discharging mode of the circuit. - i (R2 + R3) = 0 R2 + R3 = R' = 4+3 = 7 kilo ohm = 7x 10 ohm -7 - iR' = 0 - - R4 = 0 - = R' 9= 9o euCK CR' is the time cons tan t for this circuit when the switch is at the position 3 7 = CR = 10 x 10-6 x 7x 103 = 0.07 s Answer : The time cons tan t when the switch is in position 3 is 0.07 s Step 3 The voltage is to be determined when the position of the switch is at 1 from the equation (1), we have During the charging mode (:) = 0 = v (1 -eCK) v(:) v(): = cV (1 -e CK) = v (1 - euCR) = 100 x (1- e.0s) Att = 60 ms = 60 x 103 s = 0, 06 v(:) = 100 x (1-e a) = 100 x 0. 6988 = 69. 88 V Answer : The voltage across the capacitor at time t = 60 ms is 69. 88 V
Transcribed Image Text:More than three subparts were required to be answered, so the 1st three has been answered. From the working principle of a capacitor it is understood that the capacitor offers zero resistance when it is uncharged and in that condition it can be taken as short circuit. When the capacitor is fully charged it will offer infinite resistance. When the position of the switch is at position 1; the capacitor will be charging from a 100 V battery; Applying Kirchhoff's Voltage law - - i (R, + R2) + V = 0 R = 2 kilo ohm, R2 = 3 kilo ohm and R3 = 4 kilo ohm R = R + R2 = 2+3 = 5 kilo ohm = 5 x 10 ohm V = 100 V The capacitor C = 10 µF -- iR + V = o i = 4 -온-R + V=0 V - = R di Now integrating the above equation, we get () q = CV (1 - e CR) CR = T = time cons tan t = 10 x 10-6 x 5x 10 = 0.05 s Answer : The cons tan t of the circuit when it is at position 1 is 0. 05 s Step 2 When the switch is at position 3, we again apply the Kirchhoff's Voltage law to the circuit, This is the discharging mode of the circuit. - i (R2 + R3) = 0 R2 + R3 = R' = 4+3 = 7 kilo ohm = 7x 10 ohm -7 - iR' = 0 - - R4 = 0 - = R' 9= 9o euCK CR' is the time cons tan t for this circuit when the switch is at the position 3 7 = CR = 10 x 10-6 x 7x 103 = 0.07 s Answer : The time cons tan t when the switch is in position 3 is 0.07 s Step 3 The voltage is to be determined when the position of the switch is at 1 from the equation (1), we have During the charging mode (:) = 0 = v (1 -eCK) v(:) v(): = cV (1 -e CK) = v (1 - euCR) = 100 x (1- e.0s) Att = 60 ms = 60 x 103 s = 0, 06 v(:) = 100 x (1-e a) = 100 x 0. 6988 = 69. 88 V Answer : The voltage across the capacitor at time t = 60 ms is 69. 88 V
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