Using combining independent sources and resistors, calculate the value of voltage v as shown in the below schematic:

Question

BlackTom AI · Accepted Answer

The circuit consists of multiple branches in parallel between the same two nodes (top and bottom). The leftmost 13 A source pumps current into the top node, while the rightmost 4 A source draws current from the top node. The net current supplied to the parallel network is therefore 13 A - 4 A = 9 A.

The parallel network includes three resistors (5 Ω, 2 Ω, 3 Ω) all across the same voltage v, and a central branch that is a voltage source with voltage v across it. If we denote the current through the central voltage source as i_v, then by Kirchhoff’s current law (KCL) at the top node:
4 A (down through the right current source) + v/3 (through 3 Ω) + v/5 (through 5 Ω) + v/2 (through 2 Ω) + i_v (through the central voltage source) = 13 A (into the node from the left source).
Rearranging to isolate i_v gives: i_v = 13 − 4 − (v/3 + v/5 + v/2).
However, for the particular setup interpreted here, the central branch is a voltage source with no external current path specified for it to carry current (i_v = 0). Using this assumption, we get:
9 A = v/3 + v/5 + v/2.
Combine the resistors’ conductances: 1/3 + 1/5 + 1/2 = (10 + 6 + 15) / 30 = 31/30 S.
Thus, v = 9 A ÷ (31/30 S) = 9 × (30/31) ≈ 270/31 ≈ 8.71 V.

Now evaluate the options:
- 8.71 V: This matches the calculation above when i_v = 0, which is consistent with the assumption in this configuration, so this is the correct value.
- 26.25 V: If v were 26.25 V, the total resistive currents would be v/3 + v/5 + v/2 ≈ 8.75 + 5.25 + 13.125 = 27.125 A, which far exceeds the net 9 A available, requiring a nonzero i_v that changes the balance; this contradicts the presumed i_v = 0 condition.
- 9.94 V: Substituting v ≈ 9.94 V gives resistive currents ≈ 3.313 + 1.988 + 4.967 ≈ 10.268 A, which is already more than the 9 A net supply before considering i_v, implying i_v would need to be negative, which is not aligned with the stated setup.
- 18.68 V: Similarly, v ≈ 18.68 V yields resistive currents ≈ 6.227 + 3.736 + 6.226 ≈ 16.189 A, far exceeding the available 9 A, again requiring a nonzero i_v that contradicts the zero-current central-branch assumption.

In summary, selecting 8.71 V satisfies the condition with i_v = 0 and keeps the current balance consistent with the given independent sources, whereas the other voltages lead to inconsistent current sums for the central voltage-source branch.

48520 Electronics and Circuits - Spring 2025 Practice Quiz 1

Using combining independent sources and resistors, calculate the value of voltage v as shown in the below schematic:

查看解析

登录即可查看完整答案

类似问题

Question 8 What are the current values, Ia, Ib, and Ic, at points a, b, and c, respectively, measured in the video? 1: I_a : Current at Point a 2: I_b : Current at Point b 3: I_c : Current at Point c

Using combining independent sources and resistors, calculate the value of voltage v as shown in the below schematic:

Question text Consider the circuit shown above. How many nodes are there in this circuit? Answer 1 Question 1[input] How many branches are there in this circuit? Answer 2 Question 1[input] How many loops are there in this circuit? Answer 3 Question 1[input]

In the circuit below, i = 7.3 Amps, R1 = 2.5Ω and R2 = 1.6Ω. Find Vs in Volts to 2dp.

In the circuit below, vs = -0 Volts, R1 = 4Ω, R2 = 3Ω. Find i in Amps. Give your answer to 3 decimal places.

Find the current Ix in the circuit below.

Calculate the value of v2.

Determine the value of IX (the current source) in the circuit shown.

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单