What Happens to the Brightness of Bulb D When the Switch is Closed?

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    Circuits Resistance
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Discussion Overview

The discussion revolves around the effects on the brightness of bulbs A and D when a switch in an electrical circuit is closed. Participants explore concepts related to current, resistance, and the configuration of the circuit, including series and parallel connections. The conversation includes theoretical reasoning and calculations regarding the behavior of the circuit before and after the switch is closed.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that closing the switch will decrease the brightness of bulb D due to current splitting between D and the parallel combination of bulbs B and C.
  • Others propose that the overall brightness of bulbs may increase because the equivalent resistance decreases when the switch is closed, allowing more current to flow.
  • One participant assigns specific resistance and voltage values to analyze the current and voltage drop across each bulb, questioning whether others can perform similar calculations.
  • Some participants emphasize that the brightness of a bulb is determined by current intensity and resistance, raising questions about how these factors interact.
  • There are conflicting views on whether bulb E's brightness changes when the switch is closed, with some asserting it remains constant due to its independent pathway.
  • Several participants express confusion over the calculations and the relationships between current, resistance, and brightness, indicating a need for clarification.
  • One participant mentions that when resistors are connected in parallel, the equivalent resistance decreases, which affects the current distribution in the circuit.
  • Another participant points out that A and D are in series, meaning their combined resistance affects the current flowing through them.
  • Some participants assert that the total current is irrelevant for determining the brightness of individual bulbs, while others challenge this assertion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the effects on the brightness of bulbs A and D when the switch is closed. Multiple competing views and calculations are presented, leading to ongoing debate and confusion regarding the relationships between current, resistance, and brightness.

Contextual Notes

Participants reference specific resistance values, voltage, and current calculations, but there are unresolved mathematical steps and assumptions that may affect the conclusions drawn. The discussion reflects varying levels of understanding of electrical circuits and the principles governing them.

  • #31
I assigned the voltage to V and the resistance of each light to R. Using V = I * R, parallel resistance equivalence, and current division I was able to algebraically determine current through the ABCD leg and then the voltage drop through A and D.

Skip intuition on this one it trips you up.
 
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  • #32
2milehi said:
Skip intuition on this one it trips you up.
I noticed that too - both the current and the voltage changes.

What I have been trying to get OP to realize is that it is the power dissipated in the bulb that determines the brightness. This is proportional to the square of the current or the inverse square of the voltage.

Since the possible answers are:
brighter,
dimmer,
stays the same
... there is a 1/3 chance of getting the right answer just by guessing.
It follows that the reasoning that is followed needs to be clear and correct to get good marks.
 

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