anonymous
  • anonymous
A 4.7 kΩ resistor dissipates 0.75 W. The voltage is
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[P=IV=V^{2}/R, V =\sqrt{PR}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
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

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anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
ok now i understand thank you so much now why couldn't my professor do this from the start lol thanks so much guys
anonymous
  • anonymous
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
anonymous
  • anonymous
The resistance in electronics is opposition of flow of electrical current.More resistance less current flow.Less resistance,more current flow.
anonymous
  • anonymous
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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