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 ๐‘ฃ๐‘ and the current ๐‘–๐‘ if the capacitor was initially uncharged and the switch is thrown into position 1 at ๐‘ก=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 ๐‘ก=10 ms (assume no leakage current). (5) E 24V 8kฮฉ 0.08ยตF R1 R2 C 20kฮฉ iC vC + - 1 2 3 Figure 2(a) E 36V 8kฮฉ 2H R1 R2 L 4kฮฉ 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 ๐‘ก=0 s. Write the mathematical expressions for ๐‘–๐ฟ , ๐‘ฃ๐ฟ , ๐‘ฃ๐‘…1 , ๐‘ฃ๐‘…2 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 ๐‘–๐ฟ , ๐‘ฃ๐ฟ , ๐‘ฃ๐‘…1 , ๐‘ฃ๐‘…2 . (5) [20]


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

Ctotal=C1C2C1+C2\frac{C_1C_2}{C_1+C_2}

=0.08โˆ—0.120.08+0.12=0.048ฮผF\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)=1820L=9mA\frac{18}{20L}=9mA

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

As capasitors are charging from t=0;

Vc(t)=Vc(โˆž)(1โˆ’eโˆ’tT)V_c(t)=V_c(\infty)(1-e^{\frac{-t}{T}})

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

Vc(t)=18(1โˆ’eโˆ’tT)VV_c(t)=18(1-e^{\frac{-t}{T}})V

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

โ€…โ€ŠโŸนโ€…โ€ŠIc(t)=0.048โˆ—10โˆ’6โˆ—ddt+18(1โˆ’eโˆ’t0.96m)A\implies I_c(t)=0.048*10^{-6}*\frac{d}{dt}+18(1-e^{\frac{-t}{0.96 m}})A

=0.048โˆ—10โˆ’6โˆ—18โˆ—(eโˆ’t0.96m0.96m)=0.048*10^{-6}*18*(\frac{e^{\frac{-t}{0.96m}}}{{0.96m}})

Ic=0.9eโˆ’t0.96mmAI_c=0.9e^{\frac{-t}{0.96m}}mA


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