Answer to Question #245721 in Mechanical Engineering for mds

Question #245721

2.1 For the circuit shown in Figure 2(a), the switch was initially at Position 2: 2.1.1 Write the mathematical expressions for the transient behavior of the voltage 𝑣𝑐 and the current 𝑖𝑐 if the capacitor was initially uncharged and the switch is thrown into position 1 at 𝑑=0 s. (5) 2.1.2 Write the mathematical expressions for the voltage and the current if the switch is moved to position 3 at 𝑑=10 ms (assume no leakage current). (5) E 24V 8kΞ© 0.08Β΅F R1 R2 C 20kΞ© iC vC + - 1 2 3 Figure 2(a) E 36V 8kΞ© 2H R1 R2 L 4kΞ© iL vL + - vR2 + - + vR1 - Figure 2(b) Figure 2: 2.2 For the circuit in Figure 2 (b): 2.2.1 The switch is closed at 𝑑=0 s. Write the mathematical expressions for 𝑖𝐿 , 𝑣𝐿 , 𝑣𝑅1 , 𝑣𝑅2 within five time constants of the storage phase. (5) 2.2.2 The switch is opened after five time constants of the storage phase, write the mathematical expressions for 𝑖𝐿 , 𝑣𝐿 , 𝑣𝑅1 , 𝑣𝑅2 . (5) [20]


1
Expert's answer
2021-10-04T02:30:11-0400

Ctotal="\\frac{C_1C_2}{C_1+C_2}"

="\\frac{0.08*0.12}{0.08+0.12}=0.048 \\mu F"

As switch is moved to position 1 at t=0, initially capacitor acts as a short circuit so Vc(0)=0v

ic(0)="\\frac{18}{20L}=9mA"

Time constant T=RC=20*0.47=0.96ms

As capasitors are charging from t=0;

"V_c(t)=V_c(\\infty)(1-e^{\\frac{-t}{T}})"

"V_c(\\infty)" is calculated at t="\\infty" when capacitor acts as an open circuit, so "V_c(\\infty)=18V"

"V_c(t)=18(1-e^{\\frac{-t}{T}})V"

"I=\\frac{cdV_c}{dt}"

"\\implies I_c(t)=0.048*10^{-6}*\\frac{d}{dt}+18(1-e^{\\frac{-t}{0.96 m}})A"

"=0.048*10^{-6}*18*(\\frac{e^{\\frac{-t}{0.96m}}}{{0.96m}})"

"I_c=0.9e^{\\frac{-t}{0.96m}}mA"


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