Finding current with parallel resistors

[cm846]

Homework Statement

R1 = 2 Ω R2 = 5 Ω R3 = 11 Ω R4 = 10 Ω V = 7 V

What is the current through R2?

Homework Equations

Req for parallel resistors: 1/Req = 1/R1 + 1/R2...
Req for series resitors: Req = R1 + R2...
Electric potential V = current I x resistance R
I have found that Req is 1.755 and the current supplied by the battery is 3.988.

The Attempt at a Solution

I tried taking the 3.988 x 3.33 (the Req for R2 and R4) which is 13.29V and since the voltage drop is the same for parallel resistors I then divided by the 5 ohms to get 2.658 amps but that's not right.

 

[tiny-tim]

Welcome to PF!

R1 = 2 Ω R2 = 5 Ω R3 = 11 Ω R4 = 10 Ω V = 7 V
…
I have found that Req is 1.755 and the current supplied by the battery is 3.988.

Hi cm846! Welcome to PF!

How did you get the 3.988?

 

[cm846]

I just took the 7V divided by the Req of 1.755 to get 3.988

 

[tiny-tim]

But that would require the battery and the R2-R4 combination to be the only things in the loop.

 

[cm846]

3.99 is the answer to the question "What is the current supplied by the battery?" so I guess I'm not sure how I'd find it any other way than the way I showed?

 

[tiny-tim]

Sorry, I mistook the 1.755 it for the R2-R4 Req.

I should have asked, where did the 1.755 come from?

 

[cm846]

Thats ok. I got the 1.755 for the Req for the entire circuit: (R2) 1/5 + (R4) 1/10 = 1/Req which is 3.33 then I added R3. 11 + 3.33 = 14.33 then 1/14.33 + (R1) 1/2 = 1/Req which is 1.755.

 

[tiny-tim]

ah, got it …

that's the current through the battery itself …

but you need the current through the loop containing the battery and R2 R3 and R4.