I'm confused as how to find the current for a circuit with both parallel and series resistors?

Question

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

R2 and R4 are parallel, then these are in series with R3, then this whole system is in parallel with R1.

I need to find the current through R2. I have found that Req for the whole circuit is 1.755 and the current supplied by the battery is 3.988. I tried to then multiply 3.988 by 3.333 (which is the Req for R2 and R4) to find the voltage drop, 13.29 then divided by R2 of 5 ohms, getting an answer of 2.658.

I have no idea if my strategy is even correct.

Answer 1

FYI - The symbols for ohms didn't work, that's what the U's with accents are.

Answer 2

R2,R4 parallel is 5*10/15=3.33 ohm
in series with R3>>>14.33 ohm
So if 7 volet is across, then half amp through the circuit, it divdes so that 2/3 goes through the five ohm resistor, or I2=2*.5amp/3=1/3 amp. Check my math.

Answer 3

R2 and R4 first
1/Req = 1/5 +1/10 = 3/10
so Req =10/3
add that to R3
10/3 +11 = 43/3
now
1/Rtotal = 1/2 + 3/43 = 49/86
so
Rtotal = 86/49 = 1.755 check
so
i total = 7/1.755 = 3.988 check
However that does not interst me because I want the R2 current
current through R2-R3-R4 branch =
7/(43/3) = 21/43 = .488 amps
voltage drop across R3 = 11*.488
= 5.37 volts
so voltage drop across R2 =7-5.37 = 1.63 volts
so
i = 1.63/5 = .325 amps

Now, when you got that 13.29 volts, you knew something was wrong because it could never be bigger than the 7 volt battery.

Answer 4

thanks for your help!!