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Homework Help: What is the fraction of the total power wasted in the leads?

  1. Sep 6, 2005 #1
    Eight lights are connected to in parallel to a 110V power source by two leads of total resistance 1.5 ohm. If 240 mA flows through each lamp, what is the fraction of the total power wasted in the leads?

    You got a 6 Volt battery, and have to design a voltage divider using 1 ohm resistors to achive an voltage of 4 volts. I get the answer to be three with trial and error, but there must be a proper approach to the question?

    A 2.8 kiloohm and a 2.1 kiloohm resistor are connected parallel, this combination is connected in series with a 1.8 kiloohm resistor. if each resistor is rated at 0.5 W, what is the maximum voltage that can be passed across the network?

    I get the combined resistance to 3 kiloohms. But then using R = V^2 / P i dont get the correct voltage. Why?
  2. jcsd
  3. Sep 6, 2005 #2


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    What did you find for this part?

    For an analytical approach, check here http://www.aikenamps.com/VoltageDividerRule.htm [Broken]. There is still a certain amount of trial in choosing R1 & R2, until you find integral values (divisible by one) for each.

    You are given the power rating for each resistor is 1/2 Watt
    Hint: that indirectly tells you how much current is allowed through each resistor. But remember that the first two resistors are in parallel so the current splits between them, according to Kirchoff's Rule. And the third resistor (1.8K) will carry the sum of the currents going through the parallel combination. Of the total voltage applied, some of the voltage will drop across the parallel combination and the rest will drop across the third resistor. With this information, you should be able to calculate the maximum voltage that can be passed by the network.
    Last edited by a moderator: May 2, 2017
  4. Sep 10, 2005 #3


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    The total power delivered by the power supply to the circuit is just the total current times the supply voltage. The leads act like a series resistor that needs to conduct all the current going through the lamps (it's like replacing the leads with a single series resistor and leads with no resistance). Using it's resistance and the total current will give the power dissipated in the leads.
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