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Need a problem solved

  • 10-03-2006 2:56pm
    #1
    Registered Users, Registered Users 2 Posts: 4,473 ✭✭✭


    If the circuit in figure 2 has the following known values; V = 90 volts, R2 = 15 Ω, R3 = 30 Ω and the power dissipated on R1 is 120 watts calculate the value of the current, the resistance R1, the potential drops, the total power dissipated.

    my.php?image=circuit2rl.png


Comments

  • Closed Accounts Posts: 5,064 ✭✭✭Gurgle


    Its like this:
    > V2 = V3 = V - V1 = I2 x R2 = I3 x R3
    > I1 = I2+I3
    > I1 * I1 * R1 = 120W
    Now plug in the numbers you have for R2 and R3.


  • Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    Ok, I did this really quickly, so there might be a mistake.

    Three equations are obvious

    1. 90 = I1*R1 + I3*R3
    2. 90 = I1*R1 + I2*R2
    3. I1 = I2 + I3

    From 1 and 2, it can be seen that I3*R3 = I2*R2 or that I2 = 2*I3. Hence I1 = 3*R3

    P1 = V1*I1 = ((I1)^2)*R1. Thus R1 = P1 / (I1^2)

    Sub this into 1 and combine with I1 = 3*I3 to get a quadratic.


  • Registered Users, Registered Users 2 Posts: 4,473 ✭✭✭R0ot


    Thanks much appreciated :D


  • Registered Users, Registered Users 2 Posts: 16,202 ✭✭✭✭Pherekydes


    The total power dissipated = 120 + power across the parallel network

    This leads to the same quadratic as Dudara's but a little shorter perhaps.

    Remember, you can always simplify a circuit by getting the equivalent resistance of parallel sections.


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