# CS计算机代考程序代写 Mid-term Exam Peer Review Kris Yu Hin Choy z5076450 Question 2i)

Mid-term Exam Peer Review Kris Yu Hin Choy z5076450 Question 2i)

Identify all nodes, a ground node, define the voltages 𝑣1, 𝑣2 and 𝑣𝐺 for the nodes. Define directions of current.

Apply nodal analysis at all non-reference nodes.

At node 1, the voltage 𝑣1 is defined by the voltage source:

At node 2, apply KCL:

𝑣1 =10−0=10𝑉. 0=𝑖1 +𝑖2 +𝑖3 +𝑖4 +𝑖5 +𝑖6

𝑣2 + 𝑣2 + 𝑣2 + 𝑣2 − 𝑣1 + 0.5𝑖1 + 1 = 0 5 10 20 20

Subbingin𝑖1 =𝑣2 and𝑣1 =10: 5

𝑣2 +𝑣2 +𝑣2 +𝑣2 −10+𝑣2 +1=0 5 10 20 20 10

4𝑣2 +2𝑣2 +𝑣2 +𝑣2 −10+2𝑣2 =−20 20 20 20 20 20 20

10𝑣2 = − 10 20 20

𝑣2 = −1 𝑉 Therefore, at the 10Ω resistor, using Ohm’s Law:

𝑖2=𝑣2 =−0.1𝐴 10Ω

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Question 2ii)

The dependent current source is the third element from the right.

Voltage across source 𝑣 = −1 𝑉

Current through source 𝑖 = 0.5 × 𝑣2 = − 1 𝐴 5 10

Therefore, we have:

Using the definition of power, and noticing current flows into the positive end:

Question 2iii)

The dependent current source is the third element from the right.

Voltage across source 𝑣 = 10 𝑉

Current through source 𝑖 = 𝑣2−𝑣1 = −0.55 𝐴 20

Therefore, we have:

𝑝 = 𝑣𝑖 = 1 × 0.1

= 0.1 𝐴 (absorbed)

Using the definition of power, and noticing current flows into the negative end:

𝑝 = 𝑣𝑖 = 10 × 0.55 = 5.5 𝐴 (supplied)

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Question 3

Define the directions of each mesh and the terminals for resistors between meshes. Note that Mesh 2 and 3 have a current source in between – this means that mesh 2 and 3 is a supermesh.

Apply mesh analysis to mesh 1 and the supermesh.

For mesh 1, applying KVL and going clockwise from the independent voltage source:

−3+1(𝑖1 −𝑖3)+1(𝑖1 −𝑖2)=0

Similarly for the supermesh, going clockwise from the dependent voltage source: 2𝑉 +1𝑖 +1𝑖 +2𝑖 −1(𝑖 −𝑖 )−1(𝑖 −𝑖 )+1𝑖 =0

𝑥32212133

2𝑉 −2𝑖 +4𝑖 +3𝑖 =0 𝑥123

We can use Ohm’s Law to find 𝑉𝑥:

𝑉 =1𝑖

2𝑖1 − 𝑖2 − 𝑖3 = 3 (1)

𝑥2

−2𝑖1 +6𝑖2 +3𝑖3 =0 (2)

The supermesh current function:

Therefore, the three equations (1), (2) and (3) are equations that can be used to solve mesh currents.

𝑖2 − 𝑖3 = 2 (3)

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Question 4i)

To find the Thévenin equivalent resistance, disable current sources (open circuit) and voltage sources (short circuit), and locate nodes.

The 4 kΩ and 2 kΩ resistors are short circuited – 𝑅𝑇h = 3𝑘Ω

For Thévenin equivalent voltage, find voltage difference between A and B. Nodal analysis:

At node 1:

At node 3:

𝑣1 =12−0=12 (1)

𝑖1 + 2𝑚 = 0 = 𝑣1 − 𝑣3 + 2𝑚 3𝑘

𝑣1 − 𝑣3 = −6 (2)

Using Equation (1) into (2):

12−𝑣3 =−6→𝑣3 =18𝑉 (2)

Therefore, 𝑉𝑇h = 18 𝑉 and 𝑅𝑇h = 3 𝑘Ω

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Question 4ii)

The equivalent circuit is on the right. Using KVL:

−18 + 6𝑘𝑖 = 0

Question 4iii)

Superposition: for each independent source, turn off all but one and find current contribution of that source. Then, sum all current through the 3 kΩ resistor.

For the 12 V voltage source, open circuit current source and do mesh analysis at mesh 1:

−12+3𝑘(𝑖1 −𝑖2)+3𝑘(𝑖1 −𝑖2)=0

𝑖1 −𝑖2 =𝑖12𝑉 =2𝑚𝐴 (1)

For the 2 mA current source, short circuit the voltage source and do nodal analysis at node 1:

𝑖 = 3𝑚𝐴

2𝑚 = 𝑣1 + 𝑣1 3𝑘 3𝑘

𝑣1 = 3 𝑉

𝑖2𝑚𝐴 =3=1𝑚𝐴 3

(3)

Summing all currents through the 3 kΩ resistor from the independent sources using Equation (1) and (2):

𝑖 = 2 + 1 = 3 𝑚𝐴

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Question 5i)

At 𝒕 = 𝟎− (steady state when switch is open):

Using KVL, going around clockwise:

−10 + 10𝑖 + 40𝑖 + 50𝑖 + 35 = 0 𝑖 = −0.25

𝑉𝐶 − 10 = 0.25 10

Because voltage does not change instantaneously:

𝑉𝐶(0−) = 𝑉𝐶(0+) = 12.5 𝑉 At 𝒕 = ∞ (steady state when switch is closed):

Because the same current flows through the 40 Ω and 10 Ω resistors (nodal analysis): 0 − 𝑉𝐶 = 𝑉𝐶 − 10

40 10 −𝑉𝐶 = 4𝑉𝐶 − 40

𝑉𝐶 = 8 𝑉

The Thevenin resistance connected to the capacitor is (40||10) = 8 Ω and therefore the time constant 𝜏 = 𝑅𝐶 = 8 × 1𝑚 = 0.008 𝑠

∴ 𝑣𝑐(𝑡) = 𝑣(∞) + [𝑣(0) − 𝑣(∞)]𝑒−𝜏𝑡

−𝑡 𝑣𝑐(𝑡)=8+4.5𝑒 0.008𝑉,𝑡→𝑠

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Question 5ii)

Question 5iii)

𝑖𝑐(𝑡) = 𝐶 𝑑𝑣 𝑑𝑡

−4.5 − 𝑡 𝑖𝑐(𝑡)=0.001×0.008𝑒 0.008

−𝑡

Because we know 𝑉𝑐 changes at the rate 8 + 4.5𝑒 0.008 𝑉, and the opposite side of the resistor is our

reference voltage 𝑉𝐺 = 0:

−𝑡 𝑖𝑐(𝑡)=−0.5625𝑒 0.008𝐴,𝑡→𝑠

𝑣𝐶−01 −𝑡 𝑖𝑅40(𝑡) = 40 = 40 (8 + 4.5𝑒 0.008 𝑉)

−𝑡 𝑖𝑅40(𝑡)=0.2+0.1125𝑒 0.008𝐴,𝑡→𝑠

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