two resistances of 28 and 25 ohms are connected in parallel;in series with this combination is a 38-?

Question

ohm reistance; and in parallel with this total combination is a 24 ohm resistance. given a voltage of 1.23volts , the total
current flowing through the circuit is __________milliamps
i neeed help whats the answer

Answer 1

ok first need to combine the resistors then use ohms law

resistors in parallel are 1/Req = 1/R1 + 1/R2
so 1/Req = 1/28 + 1/25...
1/Req = 0.0357 ohms +0.04ohms
Req = 13.21 ohms

now add to series resistor:

Req = R1 + R2.... Req = 38 + 13.21 = 51.21 ohms

again add this in parallel to 24 ohms resistor:

1/Req = 1/51.21 + 1/24 = .0195+ .04167 = .06116 ohms
Req = 16.35 ohms

ohms law: V = IR
therefore I = V/R= 1.23/16.35=0.0752 amps

to get milliamps divide by 1000 so answer is:
0.0000752 milliamps
or 7.52 * 10^-5 Amps

Answer 2

This is the correct answer and equation: R1 = 28 , R2 = 25
R3 = R1*R2 / R1+R2
R3 = 13.20 ohms
Given R4 = resistance in series = 38 ohms
Final resistance R= R3+R4=38+13.20=51.20 ohms
now voltage = current*resistance
or V=IR where V=voltage,I=current and R= resistance
from this we get I = V/R
given V=1.23
therefore I=1.23/51.20=0.02402 amperes
and 1 ampere = 1000 milliampere
so I=0.02402*1000=24.02mA

final answer is: 24.02mA

Oh and that guy, cgsna does not know what he's talking about! While it's true the other two people got it wrong...my answer is correct! At least they attempted to answer, the other guy, cgsna, didn't even bother, and yet he puts others down for trying.

Answer 3

1.23/(1/(1/(1/28 + 1/25) + 38) + 1/24)

Do the math yourself. I can't believe you got so many stupid, incorrect, answers.

Looking back, I take that back. Tycha, yours looks correct (without doing the actual math), sorry!

Answer 4

First 1/(1/28 + 1/25) = 13.2 ohm

13.2 + 38 = 51.2 ohm

1/(1/51.2 + 1/24) = 16.3 ohm

1.23 / 16.3 = 0.0752 amps = 75.2 mA.