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

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When the switch is closed, the brightness of bulb D decreases due to its parallel connection with bulbs B and C, which lowers the overall resistance and increases current flow through the circuit. Bulb A's brightness increases as it is in series with D, but the current is shared between them, resulting in less current through A compared to when the switch is open. Bulb E remains unaffected because it is in a separate parallel pathway, maintaining its voltage and current at 1 amp. The overall decrease in resistance leads to a higher total current, but the distribution affects individual bulb brightness. Understanding the relationship between current, resistance, and power is crucial for determining bulb brightness in this circuit scenario.
  • #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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