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radar

  • one year ago

Electricity problem from Jan. 1959 issue of QST; Using Ohm's law find.

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  1. radar
    • one year ago
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    "There are three (3) resistors in series, but you only know three things: R1 has a value of 2 Ohms, R2 dissipates 2 Watts, and R3 has a voltage drop of 2 volts across it. Given that the power supply is a 10.5 Volt battery, how can you puzzle out the value of R2 and R3, as well as the current, I?

  2. radar
    • one year ago
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    Hint: Don't be surprised that there may be more than one solution.

  3. anonymous
    • one year ago
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    \[I ^{2}*R2=P2=2wats,so I=\sqrt{P2/R2}\] ,where I is current,P2 is power dissipation of R2 resistor,R2 resistane of R2 resistor,Ubattery=U3+I*R1+I*R2,where Ubattery is 10.5 U3 is 2V,R1 is 2 Ohms,and from there its quadratic journey to unknown yet still reachable positive solution of R2

  4. radar
    • one year ago
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    Yes it results in a quadratic thus the two solutions.

  5. radar
    • one year ago
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    |dw:1440505617227:dw|

  6. radar
    • one year ago
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    |dw:1440593442334:dw||dw:1440593505020:dw|Multiply all terms by I getting: \[10.5I = 2I ^{2}+2+2I\]Consolidate similar terms and place in standard form. Getting:\[2I ^{2}-8.5I+2=0\]divide thru by 2 getting the quadratic in final form: \[I ^{2}-4.25I+1=0\] Solve for the two real values of current and the values of R2 and R3 with each value of current.

  7. souvik
    • one year ago
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    .

  8. radar
    • one year ago
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    Solving the quadratic (completing the square" method Arranging the equation: \[I ^{2}-4.25I = -1\]\[I ^{2}-4.25I + 4.515625 = -1 + 4.515625 = 3.515625\]\[(I-2.125)^{2}=3.515625\]\[I - 2.125=\pm \sqrt{3.515625}\]\[I=2.125\pm 1.875\]\[I = 4\]\[I=0.25\]The conditions of current are: I=4 Amps and I can equal 250 milliamps.

  9. radar
    • one year ago
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    When I = 4 Amps Then R2 =.125 Ohm and R3 = .5 Ohm When I = .25 Amps then R2= 32 Ohms and R3 = 8 Ohms

  10. radar
    • one year ago
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    |dw:1440684714597:dw||dw:1440684916591:dw|

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