Basic Power source analysis

Discussion in 'Homework Help' started by choboko, Jun 29, 2010.

  1. choboko

    Thread Starter New Member

    Jun 2, 2009
    So maybe I'm overthinking or something but here's the question:

    Power outlet open-circuit voltage is measured at exactly 240V. The fan heater(rated at 240V and 2kW) is connected to outlet and voltage V drops to 236.4V. Find the Series resistance (Rs).

    From the previous question I calculated the load resistance (Rl) to be 28.8ohms (V^2/P) - to which i checked the answers and is correct. However for Rs it's 04386 ohms and I have no idea how they got that.

    Any help would be great.
  2. tskaggs

    New Member

    Jun 17, 2010
    Have you tried figuring out the load current and working backwards?
  3. choboko

    Thread Starter New Member

    Jun 2, 2009
    Not quite sure what to do after that
  4. choboko

    Thread Starter New Member

    Jun 2, 2009
    i get a current value of (236.4/0.4386) = 538.99A...then where to go from there?
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    That won't help you.

    Consider the circuit as a 240V ideal source in series with an unknown source resistance Rs and the fan heater.

    So you have a voltage drop across the heater of 236.4V. As you state the load resistance is 28.8Ω.

    How much current flows in the 28.8Ω load with a voltage drop of 236.4V? This will give you the load current Iload which also flows in the source resistor Rs.

    You can work out the voltage drop across Rs is 240-236.4=3.6V.

    So if you know the voltage drop across Rs and the current flowing in it [=Iload] what is the resistance value of Rs?
    choboko likes this.