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

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SUMMARY

When the switch is closed in a circuit with bulbs A, D, B, and C, the brightness of each bulb changes due to variations in current and resistance. Bulb D, which is in parallel with bulbs B and C, experiences a decrease in brightness because the total current is split among them, while bulbs A and E maintain their brightness. The equivalent resistance of the circuit decreases, leading to an increase in total current, but the distribution of current results in bulb D dimming. The power dissipated in each bulb, which determines brightness, is directly related to the current flowing through it.

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  • Knowledge of Ohm's Law (V = I * R)
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  • Ability to calculate equivalent resistance in parallel circuits
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  • #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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