What is the potential drop and current in this circuit?

[wwari]

[URL]http://www.fizikodev.itu.edu.tr/res/mcgrawhill/BauerWestfall/1stEdition/Chapter25/P076figure.png[/URL]A battery with V = 2.27 V is connected to three resistors as shown in the figure, where R1 = 1.61 Ω, R2 = 4.71 Ω, and R3 = 5.07 Ω.

a) Find the potential drop across resistor 2.


b) Find the current in resistor 3.

 

[NascentOxygen]

Is there a figure to accompany this?
Do we see your attempt at the solution? If it is just 3 resistors in series, then how much current is flowing through them?

 

[wwari]

there is no problem about picture.Some part of picture is not clear.But it hasnt deficient information.It is my homework and I can't find resultant resistant.

 

[NascentOxygen]

there is no problem about picture.Some part of picture is not clear.

I don't see any picture.

 

[asteriatic]

a) The potential drop across resistor 2 can be calculated using Ohm's Law, V = IR. Thus, the potential drop across resistor 2 is V2 = (4.71 Ω)(0.48 A) = 2.26 V.

b) The current in resistor 3 can be calculated using Ohm's Law, I = V/R. Thus, the current in resistor 3 is I3 = (2.27 V)/(5.07 Ω) = 0.45 A.