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Some electricy probs

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    You are given a 120 V battery; a hair dryer designed to work properly at 120V, where its rated at 1200 W; and a mixer designed to work properly at 60v, where its rated at 120W. You also happen to have a supply of 60 ohm resistors.

    a) determine the resistance of the hairdryer and the mixer at their rated voltages

    B) what current is established in the mixer when its working properly

    C) using the symbols shown, draw the connections needed to make both devices work simultaneously
    a resistor
    a battery
    a box with the letter h signfiying hair dryer
    and a box signifying mixer

    d) what power must the battery supply to run your circuit?
    2. Relevant equations

    3. The attempt at a solution
    a)I think the resistance of the hairdryer is 12, but how would I found the mixers?
    B) dont know how to approach this one?
    c) same as above
    d) wait wait... nope, dont know how to this one either.

    My teacher likes to assign homework on stuff we havant learned yet :(
  2. jcsd
  3. Mar 1, 2009 #2
    How did you find out the resistance of the dryer? Is there an equation for power in electrics?
    Last edited: Mar 1, 2009
  4. Mar 3, 2009 #3
    Power = VI = I2R = V2 / R

    A) The hairdryer's internal resistance is correct. The mixer's internal resistance can be found the same way.

    B) When the mixer is working properly you know the voltage across its terminals is 60 V and you know how much power it is consuming, 120 W. You found the mixer's internal resistance in part A. See the above power relationships.

    C) You know the voltages both of the devices need to operate at, and you know the source voltage, 120 V. Also, you have 60 ohm resistors to use to cause voltage drops in parts of the circuit. Think about parallel and series connections.

    D) Once you have designed the correct circuit, you just need to come up with an Req and determine the source current. Then see the above power relationships.
  5. Mar 3, 2009 #4
    thanks, solved this one easy. Just needed the right push.
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