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Lab Procedure: Part 1: Source Free RC Circuit V(t) ilts C3 (R3 21ko 1uF IC=10V a) For the circuit shown above, provide the eq
Lab Procedure: Part 2: Voltage Step Response of a RC Circuit tu a) For the circuit shown above a Step voltage source, provide
Lab Procedure: Part 3: Voltage Pulse Response of a RC Circuit PULSE VOLTAGE A 1kg AC analysis magnitude AC analyse . OV 10V w
Lab Procedure: Part 4: Transient Response of a RC Circuit 3kg Ev 32 R2 24v s >5kg ka 10ms 20ms C1 1pF a) For the circuit show
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Answer #1

Design2 - Multisim - [Design2*] Eile Edit View Place MCU Simulate Transfer Tools Reports Options Window Help B BW 6 % @ 7.9 m

- a x Design2 - Multisim - [Design2 *] Eile Edit View Place MCU Simulate Transfer Tools Reports Options Window Help BBW 8 % @

- a x Transient Design2 - Multisim - [Design2 *] Eile Edit View Place MCU Simulate to Analyses and Simulation O M 6 % E é Act

choose user defined under innitial conditions to obtain the result

- a x Transient Design2 - Multisim - [Design2 *] Eile Edit View Place MCU Simulate to Analyses and Simulation O M 6 % E é Act

+ Ve(t) Apply KUL to the loop . tic R t Velt)=0 i= c duc de Educrt Vc=0 dt ur CR dve at Integrate on Both sides ren (WC) = 4-love = t - ln(10) RC In (10) -en (ve) = 7 RC hude . - ERC Vc = 10e -t/t Ve=10 ett 1. The 10e Ve(t) at t=t= = loe = 3.6tv ic- a x tho te QQ@g Design1 - Multisim - [Design1 *] Eile Edit View Place MCU Simulate Transfer Tools Reports Options Window He

27 pie ta T VE Apply Kul to the loop. avtickituc=0 ie= c dve -VHC dvert VC =0 at erdve zur Ve dt due = dt cen (v=vc) = 5 tk >at t=0 =0 -en (u)=0tk substitute in ② RC -In (v=ve) ed en los sen(u) – en [v-ve) en lu ve ) te X CRC V-vc-en -t/Rc V-Vc e -E/b) Velt) at t=20 = uci-e 2%) = uci-e?) = 0.864 V. = 0.864X10 = 8a64 volts. at t=lms, t=RC = 1kxlu = ims ve) = 10 ,-e tante) c

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