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anonymous

  • 5 years ago

A 4.7 kΩ resistor dissipates 0.75 W. The voltage is

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  1. anonymous
    • 5 years ago
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    \[P=IV=V^{2}/R, V =\sqrt{PR}\]

  2. anonymous
    • 5 years ago
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    I am still confused because my professor didnt really teach us all of this and its also not in the book that we have so can you please help me a little more

  3. anonymous
    • 5 years ago
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    W is watts, or power. 4.7kOhms is resistance. using the relationships above, you just plug the numbers into the right spot and you have your answer

  4. anonymous
    • 5 years ago
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    P=ExI=ExE/R=E^2/R P=E^2/R 0.75W=E^2/4700 ohm E^2=0.75x4700 ohm E^2=3525 E=59.3 V

  5. anonymous
    • 5 years ago
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    oh, and \[IV=V ^{2}/R\]by using I=V/R from V=IR (where I is current in amps) remember 4.7kOhms is 4700 Ohms

  6. anonymous
    • 5 years ago
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    ok now i understand thank you so much now why couldn't my professor do this from the start lol thanks so much guys

  7. anonymous
    • 5 years ago
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    A parallel circuit contains four branches with resistors values of 4.7k, 5.6k, 8.1k and 10k. Which branch has the largest current flow

  8. anonymous
    • 5 years ago
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    The resistance in electronics is opposition of flow of electrical current.More resistance less current flow.Less resistance,more current flow.

  9. anonymous
    • 5 years ago
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    To expand, in a parallel setup, the voltage drop across each resistor is the same, so you just employ V=IR. Smaller R means larger I

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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