Another Ohm's Law question that stumped me

2011 April 7
by roma

I also couldn't figure out the reasoning behind another question that pertains to the same diagram.

Page 65, question 6:

Suppose that the original timer in the figure, which has no resistance, is replaced by another timer that has resistance equal to that of the lamp. If nothing else in the circuit changes, how will the energy consumed by the lamp in 2 h be different when the switch is closed, as compared with the energy consumed by the lamp in 2 h in the original circuit when the switch is closed?

a. It will be the same.
b. It will be half as great.
c. It will be 1/4 as great.
d. More information is needed to answer this question.

Power is the product of voltage (E) and current (I). Energy is equal to the product of power and time (I know this is a backwards way of defining energy, but I'm just following Gibilisco).

In this example, we aren't given current, so we use P = E^2/R.
Thus, W (energy) = E^2t / R

According to this formula, if the resistance doubles, the energy will fall in half. So, I chose b. However, the answer is c. I don't follow.

2 Responses leave one →
  1. roma permalink*
    April 18, 2011

    FastCompany recently featured in an article about the most innovative companies in education. I decided to try it out for my electricity woes in order to answer the latter question (and others). I registered and posted a link to this question 6 hours ago. Four hours later I received a response that resolved my misunderstanding.

    First, use Ohm's law to determine how the current in that part of the circuit will be affected.

    I = \displaystyle\frac{E}{R}

    The resistance doubles, so

    \displaystyle\frac{I}{2} = \frac{E}{2R}

    The current will be one half as great.

    \displaystyle Energy = EIt

    We're given the resistance, so we substitute IR for voltage

    \displaystyle Energy = I^2Rt

    Now, the tricky part is that we are asked about the energy consumed by the lamp ONLY.
    So while the current does change to accommodate the change in resistance, the lamp's resistance stays the same.

    Energy consumed by lamp in the original circuit:
    \displaystyle Energy = I^2Rt

    Energy consumed by the lamp in the new circuit:
     \displaystyle Energy = (\frac{I}{2})^2Rt
    = \displaystyle\frac{I^2Rt}{4}

    So, the energy consumed by the lamp in the modified circuit is a quarter of the energy consumed by it in the original circuit. So, the answer is c.

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