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Writer's picturekeshprad

PhysicsBowl 2017 Q29


29) This question asks us to find how the brightness of the bulbs changes when we move the bulbs from a series circuit to a parallel circuit. First of all, since bulb X is brighter than bulb Y when in series, we know that one bulb must have greater resistance than the other. For problems like this, it is easiest to explain using example numbers.


As a reminder, a bulb can be treated like a resistor, and the brightness of a bulb is determined by the power it dissipates.

Where P is power, I is current, and V is voltage

Let's just use example numbers and solve for the power dissipated. by each resistor.

Where P is power, I is current, R is resistance, and V is voltage

Remember that the question told us that when in series, bulb X is brighter. This means we can rename the 4 Ohm resistor, R[1], as bulb X, and R[2] as bulb Y.


Next, we can take a look at the bulbs in parallel.

Where P is power, I is current, R is resistance, and V is voltage

Answer: E

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