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In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R₁ = 11.0 kn, R₂ = 22.0k, R3 = 4.00 k2, and C= 16.0 μF.) 9.00 V (a) Find the steady-state current in each resistor. I1 = I2 = 13 = Need Help? R₁ ww μА ДА μА (b) Find the charge Qmax on the capacitor. μC R₂ ms Read It (c) The switch is now opened at t = 0. Write an equation for the current in R₂ as a function of time. (Use the following as necessary: t. Do not enter units in answers. Assume the current is in microamperes, and t is in seconds.) IR₂ = C (d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. {R3

In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R₁ = 11.0 kn, R₂ = 22.0k, R3 = 4.00 k2, and C= 16.0 μF.) 9.00 V (a) Find the steady-state current in each resistor. I1 = I2 = 13 = Need Help? R₁ ww μА ДА μА (b) Find the charge Qmax on the capacitor. μC R₂ ms Read It (c) The switch is now opened at t = 0. Write an equation for the current in R₂ as a function of time. (Use the following as necessary: t. Do not enter units in answers. Assume the current is in microamperes, and t is in seconds.) IR₂ = C (d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. {R3

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R₁ = 11.0 kn, R₂ = 22.0k, R3 = 4.00 k2, and C = 16.0 μF.) 9.00 V (a) Find the steady-state current in each resistor. I1 = I2 = 13 = Need Help? R₁ ww μА ДА μА (b) Find the charge Qmax on the capacitor. μC R₂ ms Read It (c) The switch is now opened at t = 0. Write an equation for the current in R₂ as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.) IR₂ = C (d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. {R3
Transcribed Image Text:In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R₁ = 11.0 kn, R₂ = 22.0k, R3 = 4.00 k2, and C = 16.0 μF.) 9.00 V (a) Find the steady-state current in each resistor. I1 = I2 = 13 = Need Help? R₁ ww μА ДА μА (b) Find the charge Qmax on the capacitor. μC R₂ ms Read It (c) The switch is now opened at t = 0. Write an equation for the current in R₂ as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.) IR₂ = C (d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. {R3
Step 1: What is capacitor A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. The plates can be made of various conducting materials, and the dielectric can be made of materials like ceramic, plastic, or electrolyte. "Since you have asked multipart questions, we will solve the first three question for you. If you want any specific question to be solved, the please specify the number of that question" Step 2: Current values in steady state In steady state the capacitor act as open circuit. The modified circuit in staedy circuit is drawn below T 1=9/(11K+22K) =272.72μA 9V скл MUL The current through resistor R3 is OA as it is open circuit. The resistor R₁ and R₂ are connected in series with voltage source 9V. The current through R₁ and R₂ are equal the current value is calculated as below 22125 Step 3: Current equation is evaluated In steady state the voltage across capacitor is voltage across resistor R₂ The maximum charge on capacitor is calculated as below Qmax=CV =CIR₂ =16x10(-6) x272. 72x10(-6) x 22x10(3) =95.99 C When the switch is open the reduced circuit is drawn below ли 9V 22102 I=l'e(-t/T) The time constant T is calculated as below. T=CX(R₂+R3) =16x10(-6) (22+4) x10³ =.416sec The current l' is evaluate as follow I'=Vc/(R₂+R3) =(272.72x10(-6) x22x10³)/(22K+4K) =230.7μA The current equation is as below 1=230.7e(-t/.416) μA 11k2 The capacitor will discharge through Resistor R₂ and R3. The current through R₂ is given by ики 16 4F T икъ 1254k2
Transcribed Image Text:Step 1: What is capacitor A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. The plates can be made of various conducting materials, and the dielectric can be made of materials like ceramic, plastic, or electrolyte. "Since you have asked multipart questions, we will solve the first three question for you. If you want any specific question to be solved, the please specify the number of that question" Step 2: Current values in steady state In steady state the capacitor act as open circuit. The modified circuit in staedy circuit is drawn below T 1=9/(11K+22K) =272.72μA 9V скл MUL The current through resistor R3 is OA as it is open circuit. The resistor R₁ and R₂ are connected in series with voltage source 9V. The current through R₁ and R₂ are equal the current value is calculated as below 22125 Step 3: Current equation is evaluated In steady state the voltage across capacitor is voltage across resistor R₂ The maximum charge on capacitor is calculated as below Qmax=CV =CIR₂ =16x10(-6) x272. 72x10(-6) x 22x10(3) =95.99 C When the switch is open the reduced circuit is drawn below ли 9V 22102 I=l'e(-t/T) The time constant T is calculated as below. T=CX(R₂+R3) =16x10(-6) (22+4) x10³ =.416sec The current l' is evaluate as follow I'=Vc/(R₂+R3) =(272.72x10(-6) x22x10³)/(22K+4K) =230.7μA The current equation is as below 1=230.7e(-t/.416) μA 11k2 The capacitor will discharge through Resistor R₂ and R3. The current through R₂ is given by ики 16 4F T икъ 1254k2
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