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:

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