R₁ (s) + + 4 1 S+2 10 10 + R₂(s) + -7 S (, (s) 1 S+4 S+3 C₂(5) -S
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Applying superposition or Mason Foury method, obtain the 4 transfer functions from this block diagram.
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- For the circuit shown in figure Req is: * 140 R3 R, R: 22 Rea Ru RC RS R7 33 6 ohm 24.4 ohm 34,4 ohm 44.4 ohmQ 1. Find the unit-step response for the system sown in the figure. RS). T= 1Sec S(stI) 1/5For the circuit shown in Figure Q2, determine the followi (a) The current in the inductors L1 and L2. (b) The voltage across the capacitors Cl and C2. (c) The total energy stored in the circuit. (d) The total power supplied by the source. 30V II RI -000-m 20mH 1052 50mH 2 R2 2002 300µF -000 30ml 91 300µF CI 600 μF R3 30Ω C3
- what are the complex powers of all elements in the circuit?(Sj2π,S-j1π,S2π,S1π,S6Z0*V)(surtch ) Consider the Circuit below: RI=200r a 7. R2 35mH 1000 For parts a-c assume the switch S is in position "a" after it was off for a long time. a) Find the time constant t for the switch in position "a". b) Determine the maximum inductor current, Imax. c) When will the inductor current be 95% of Imax. For parts d-f assume the switch S is in position “b" after it was in position "a" for a long time d) Find the time constant t for the switch in position "b". e) When will the inductor current be 5% of its initial value (initial value is when the switch is switched from “a" to “b") f) How much energy is dissipated by the resistance if the switch is in position "b" for a long time. eeleFor the circuit in Figure 3 the switch is in the left position for several minutes: (a) Find the Initlal voltage, V, on the capacitor just before the switch is flipped (b) Find an expression v(t) that describes the voltage across the 20 N resistor after the switch has been Figure 3 U 09 flipped to the right NOTE: Remember what we said in class: use a Circuit-Specific Equation to get a value you know. Then solve for whatever else the problem asks for +50 µF 380 0 20 2
- In the figure you have a RL circuit with direct current. All resistors have resistance 30.0 , the inductors is 3.0 H. The battery provides a voltage of 840.0 V. At t=0 close the switch. At that instant, what are a) the current in the resistor R1? b) the current in the resistor R3? c) the voltage at the inductor L? +1 V m R1 switch M R2 A A V www R3For the circuit below with input voltage V1, a step function of magnitude 25 volts at time t-0, find the transient response voltage across the capacitor C1 and plot the results: (Provide your calculations and reasoning for your answer.) R1 555 250mH V1 = 0 V2 - 25 TD = 0 TR In V1 C1 TF in PW- 25 PER = 3.3uFor the circuit below with switch U1 closing at time t=0 and the initial current in inductor L1 equal zero, find the transient response voltage across the resistor R1: (Provide your calculations and reasoning for your answer.) L1 U1 50mH 11 R2 1k R1 0.150Ado 15
- Tutorial exercise Jimmy the Circuit Builder is at it again and this time he is using AC. He has all of the linear components in pairs: two AC voltage sources, two AC current sources, two resistors, two capacitors and two inductors. He is trying to build circuits to create certain voltages and currents. All of the sources operate at w=500 rad/s. 940° V 2 <90° A - 12<60° V 1-30° A (c) V=6490° V (d) I=64150° A (e) I = 0.6 +j1.04 (f) V=18.25434.7° V + (a) Draw the circuit. (b) Calculate the voltage across each component. 100μF TE 1mF 46 392 -M 1mH hint: try combining the two sources 10mH Jimmy first builds one circuit to check his understanding of AC voltage, current and power. The circuit is a series connection of the 12 <60° V voltage source, the 3 resistor, the 10 mH inductor, and the 1mF capacitor. 1092 M Using only two elements, connect the pair of elements to create the following AC voltage and currents: Using exactly three elements, draw circuits to create the following AC voltage…Assignment: RL & RC Circuits. R Figure 1: RL Circuit 1. In an RL series circuit containing only a resistor and inductor, Kirchhop's second law states that the sum of the voltage drop across the inductor(Ldi/dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. The current i(t), called the response of the system, is governed by a differential equation di L + Ri = E(t) 'dt where L and R are constants known as inductance and the resistance respectively. Based on the above equation, solve the following: (a) A 12-volt battery is connected to a series in which the inductance is 0.5 henry and the resistance is 10 ohms. Determine the current i if the initial current is zero. Then determine the steady-state current for the system. Ans: i(t) = (1 –e-201) %3D (b) An electromotive force 0 20 120, E(t) = { 0, is applied to an LR series circuit in which the inductance is 20 henry and the resistance is 2 ohms. Find the current i(t) if i(0) = 0.…Given That: Is=0.0098, R1=60000, R2=6000 0, L=16 H, C=2 F, The circuit shown below under dc conditions find the following: L ell Is R1 R2 the current in L The voltage across capacitor