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Circuit that is both parallel and in series

  1. Jul 23, 2006 #1
    Determine the power dissipated in the 9 ohm resistor in the circuit shown in the drawing. (R1=4.0 Ohms, R2=9.0 Ohms, and V1=9V).


    When trying to solve this problem, I tired to add the resistance of the bottom two as if they were in series, and then add that to R1 to get the total resistance. I then found the current using the given voltage, and then tried to multiply that to the voltage to get an answer of 9 watts. But that is wrong, and I am not sure where I went wrong.

    I'd appreciate any help. Thank you.
  2. jcsd
  3. Jul 24, 2006 #2

    Andrew Mason

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    I can't see your drawing yet but I expect that it consists of two branches, one of which has a single 9 ohm resistor and the other with two series resistors of 4 + 9 = 13 ohms.

    For parallel connected resistors, the voltages are the same and the currents add up. In series the currents are the same and the voltages add up (to the total V).

    So the total current is the current in both branches of the parallel circuit and the current in each is V/R.

    If the applied voltage is 9 volts you will get a current of 9/9 + 9/13 amps. The power in the 9 ohm resistor is its current x voltage = 9 * 9/9 = 9 w.

  4. Jul 24, 2006 #3
    you must take all the resistor together to make 1 substitutional resistane using the following rules:

    1. serial placed resistance: [tex]R_{subst.} = \Sigma R[/tex] and U is divided over the resistors.

    2. parallel circuit: [tex]\frac {1} {R_{subst.}} = \Sigma \frac {1} {R}[/tex] and I is divided over the resistors.
    Last edited: Jul 24, 2006
  5. Jul 24, 2006 #4
    I got 9 watts as my answer, and that was incorrect. Maybe the picture that's now uploaded can help someone help me with this question? Thanks again everybody.
  6. Jul 24, 2006 #5
    what do we nee d for the power: U and I

    Now lets take the serial 1 ohm and 9 ohm and the serial 2 ihm and 1 ohm resistors together and we get a simplified circuit with a new R of 10 ohm and 3 ohm parallel to each other.

    Now we must take the parallel 10 ohm and 3 ohm toegether. this yields:

    [tex] \frac {1} {\frac {1} {10} \cdot \frac {1} {3} } = 2.31 \Omega[/tex]

    Then add these 2 resistors together to get 1 single resistor of 6.31 ohm.

    Now the source has a voltage of 9.0 V so the I through the singel resistor is [itex] I = \frac {9} {6.31} = 1.43 A[/itex]

    Thus thorugh the 2.31 resistor earlier has a current of 1.43 A with a voltage of [itex] U = 1.43 \cdot 2.31 = 3.3033 V[/itex]. (As a control you can calculate U of the 4 ohm resistor.

    Now 1 step back U is 3,3033 V over the 10 ohm resistor and 3 ohm resistor and I is divided. Over the 10 ohm resistor I is [itex] \frac {3.3033} {10} = 0.330033 A[/itex]

    Then at the original situation the current in the 9 ohm resistor is 0,33033 A and the voltage is [itex] 0.330033 \cdot 9 = 2.97 V[/itex]

    Then the power is [itex] P = 2.97 \cdot 0.33033 = 0.98 W[/itex]

    (I may have made a miscalculation --> i did not use paper)
  7. Jul 24, 2006 #6
    I have juste made the calculations and I arrived at 0.98 watt ≈ 1 watt.
    So the result of sdekivit seems ok
  8. Jul 24, 2006 #7
    I have juste made the calculations and I arrived at 0.98 watt ? 1 watt.
    So the result of sdekivit seems ok
  9. Jul 24, 2006 #8
    Also, remember the following rules:

    [tex] R_{series} = R_1 + R_2 [/tex]
    [tex] R_{parallel} = \frac{R_1 \, R_2}{R_1 +R_2} [/tex]

    Then you can use some nice short hand when you are trying to find an equivalent resistance. In this case the equivalent resistance of those resistor combinations would be:
    [tex] R_{eq} = R_1 + ((2 + 1) || (R_2 + 1)) [/tex]

    Notice that [itex] R_a|| R_b [/itex] would mean:
    [tex] R_a|| R_b = \frac{R_a \, R_b}{R_a + R_b} [/tex]
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