anonymous 5 years ago A 4.7 kΩ resistor dissipates 0.75 W. The voltage is

1. anonymous

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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. anonymous

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

9. 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