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- anonymous

A 4.7 kΩ resistor dissipates 0.75 W. The voltage is

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- anonymous

A 4.7 kΩ resistor dissipates 0.75 W. The voltage is

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- anonymous

\[P=IV=V^{2}/R, V =\sqrt{PR}\]

- 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

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

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

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

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

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

The resistance in electronics is opposition of flow of electrical current.More resistance less current flow.Less resistance,more current flow.

- 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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