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2 A 8Ω 4Ω 12 V 6Ω 6Ω HI- Figure 3.46 For Review Questions 3.1 and 3.2. 3.2 In the circuit of Fig. 3.46, applying KCL at node 2 gives: 4 1V2,V2 V 4 V1-V2 · 12-V2 V2

Question
2 A
8Ω
4Ω
12 V
6Ω
6Ω
HI-
Figure 3.46
For Review Questions 3.1 and 3.2.
3.2
In the circuit of Fig. 3.46, applying KCL at node 2
gives:
4
1V2,V2 V
4
V1-V2 · 12-V2
V2
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2 A 8Ω 4Ω 12 V 6Ω 6Ω HI- Figure 3.46 For Review Questions 3.1 and 3.2. 3.2 In the circuit of Fig. 3.46, applying KCL at node 2 gives: 4 1V2,V2 V 4 V1-V2 · 12-V2 V2

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Step 1

The purely resistive circuit provided in the question can be solved by applying Kirchoff's Voltage Law (KVL) and Kirchoff's Current Law (KCL). 

According to KCL, total current entering a node is equal to the total current leaving a node. That principle will be applied here to solve this problem. 

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Engineering

Electrical Engineering

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