Consider the circuit diagram below: Which of the following breadboard circuits implements this circuit correctly? Notes: The power supply is set to 5V. The resistors in the breadboards are all 150 Ω. This image may be useful:

Question

BlackTom AI · Accepted Answer

The schematic is a single 150 Ω resistor in series with the supply feeding two parallel branches: an LED branch and a 150 Ω resistor branch. Key points: breadboard rows form one electrical node, LED polarity matters (anode toward the positive node, cathode toward ground), and to implement a parallel branch you must place the LED and the second resistor between the same two nodes.

1) Relevant knowledge points and concepts
- Series vs. parallel: the first 150 Ω (call it R1) is in series with the supply; the LED and the second 150 Ω (R2) are each connected from the same node (node A) to the return node (node B), so they are in parallel.
- Breadboard topology: holes in a horizontal row (or vertical rail depending on the board) are electrically identical; placing components in the same row ties them together.
- LED polarity: anode must face the higher potential (node A), cathode to the lower (node B/ground).

2) Step-by-step reasoning / wiring mapping
- Identify node A: the junction immediately after R1. On the breadboard you must place R1 so its far end lands in a row that becomes node A.
- Connect R2 between node A and node B (ground row). That places R2 as the lower parallel branch.
- Place the LED with its anode in node A’s row and cathode in node B’s row so it forms the upper parallel branch.
- Tie node B to the negative terminal of the 5 V supply (ground rail). The positive supply goes to R1’s input.

3) Intermediate connection logic (no numerical currents required)
- Both LED and R2 share the same two node rows (A and B). R1 is the only element between +5 V and node A.
- This matches the schematic: one series resistor, then two branches to ground.

4) Why option d is correct
- Option d is the only breadboard image that places a resistor from +5 V into a node (R1), and then places the LED and the other 150 Ω each between that node and the ground rail (correct parallel placement), with the LED oriented correct polarity.

5) Why the other options are incorrect (briefly)
- a., b., c. each misplace components so the LED and the second resistor are not tied to the same pair of node rows (yielding wrong series/parallel relationships), or the LED is oriented or connected to the wrong rail. Some show both resistors effectively in series or the LED not sharing the correct node, so they do not replicate the schematic.

Conclusion: The correct breadboard implementation is option d. This matches the provided answer (d.).

Consider the circuit diagram below: Which of the following breadboard circuits implements this circuit correctly? Notes: The power supply is set to 5V. The resistors in the breadboards are all 150 Ω. This image may be useful:单项选择题

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