Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circuit and Power Question

  1. Jan 27, 2004 #1
    The final question of my exam review, I have some answers but I may be completely off...

    "[7] Three 100-Ohm resistors are connected as shown below. The maximum power dissipated in any one resistor is 25W. (a) What is the maximum voltage that can be applied to the terminals a and b? (b) For the voltage determined in part (a), what is the power dissipated in each resistor? (c) What is the total power dissipated?"


    For part (a) I found current 'I' by using P=R*I^2, which was 25=100*I^2, so I found 'I' to equal I=0.5A. I found the total resistance through the standard series-parallel resitance calculations, which ended up as R-Total=150 Ohms. Plugging 'I' and R into V=IR, I found V to equal 75v for part (a).

    For part (b) I used V=IR to find the individual voltage for each resistor, which ended up being 50v for the first resistor and 25v for each of the two parallel resistors. I used that value to determine the current in each resistor, which turned out to be .5A for the first resistor and .25A for the two in parallel. Using these values, I was able to plug into P=R*I^2 to determine the power amounts for each resistor, R1=25W, R2=6.25W, R3=6.25W. Is that the right process, and answers?

    In part (c) I simply used the total resistance and the series-wide current and plugged those values into P-total=R*I^2 to find P-total=150*(1/2)^2, which turned out to be P-total=37.5W. This total power value matches with the sum of the power found in each individual resistor found in part (b), which is supposed to happen.

    While the answers seem to check out, I'd like to confirm whether or not I did the problem correctly. Any opinions or suggestions?

    Thanks for your time and help.
  2. jcsd
  3. Jan 27, 2004 #2
    no fig is shown by the link
  4. Jan 27, 2004 #3

    Doc Al

    User Avatar

    Staff: Mentor

    I was able to view your picture using Netscape, but not IE.
    Looks good to me.
  5. Jan 27, 2004 #4
    using IE, I had to copy and paste the url into the address box.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook