Answer to Question #245721 in Mechanical Engineering for mds

Question

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 𝑡=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 𝑡=10 ms (assume no leakage current). (5) E 24V 8kΩ 0.08µF R1 R2 C 20kΩ iC vC + - 1 2 3 Figure 2(a) E 36V 8kΩ 2H R1 R2 L 4kΩ 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 𝑡=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]

Answer 1

C_total=$\frac{C_1C_2}{C_1+C_2}$

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

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

i_c(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 $V_c(\infty)=18V$

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

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

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

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

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