Mark K. answered • 05/22/15

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The first part of the question should be tackled by thinking about the power dissipated through the light bulbs. A good way to approach this is to determine the voltage drop across each resistor. This is the best approach because we can write the voltage in terms of the source voltage, which I am assuming is the same for each circuit. There are three equations used to find the power in a resistor:

- P = IV
- P = (I^2)R
- P = (V^2)/R

We are interested in using equation 3. For the series circuit each resistor drops half the voltage (we call this a voltage divider circuit). If we substitute the voltage drop, V/2, into equation 3.

- P = (V^2)/R <--- Equation 3.
- P = [(V/2)^2]/R <--- Substitute voltage drop across resistor from series circuit for V
- P = (1/4)(V^2)/R

Now we will do the same for the parallel circuit:

- P = (V^2)/R <--- Since the resistors are in parallel the voltage drop is equal to the source voltage. This is the power dissipated by each resistor.

We can now compare the power values from the series circuit and the parallel circuit. We observe that when the resistors are in series they dissipate 1/4 the power of the resistors in parallel. We used power to analyze the brightness since power is the rate at which energy is leaving through the resistor. The larger the power, the more light energy is being released. Therefore the parallel circuit has the brighter bulbs.