Stuck on a question

Discussion in 'Homework Help' started by cgeorge5150, Oct 8, 2009.

  1. cgeorge5150

    Thread Starter New Member

    Oct 8, 2009
    Three resistors are connected in series across 10V power supply as shown. The first resistor has a value of 1 Ohm, the second has a voltage drop of 1V, and the third has a power dissipation of 1W. Find the value of the circuit current. There are two possible answers. How do I figure this out?
    Last edited: Oct 8, 2009
  2. blueroomelectronics

    AAC Fanatic!

    Jul 22, 2007
    Post your answers.
  3. cgeorge5150

    Thread Starter New Member

    Oct 8, 2009
    i get two answers. since it is a series circuit current is constant.

    using R3 I=P/E 1W/10V=.1A
    using R2 10V/.1A= 10 Ohm
    VR2 is 1V/10 Ohm = .1A
    using R1 10V/1 Ohm is 10A
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    Why must you have two answers? Is that what the instructor stated?

    Can you tell me about Kirchoff's Voltage Law as it applies to your circuit?
    Last edited: Oct 8, 2009
  5. Ratch

    New Member

    Mar 20, 2007

    But 10 volts is not dropped only across R3, so that is wrong.

    Wrong amps from first calculation, so wrong for second calculation.

    Wrong on second calculation, so wrong on third calculation.

    Wrong because 10 volts is spread across all three resistors

    OK, lets get down to what works. Who cares what the two unknown resistors are? We are trying to find current. Looks to me like the best way to do that is to equate the power from the 10 volt source to the power dissipated by the resistors. So let's do that.

    T = total resistance of all three resistors, then I = 10/T

    The general equation is E^2/R = R*I^2 + I*V + 1

    The particular equation is 10^2/T = 1*(10/T)^2 + 1*10/T + 1

    Solving the quadratic gives T = 1.125178063, and T = 88.87482194

    So I = 10/1.125178063 = 8.887482194 amps
    and I = 10/88.87482194 = 0.1125178063 amps