Series and Parallel resistors - Computation?

Question

Okay here it goes....I am having a brain lock right here......I am in school for electronics / computer IT training one of the questions on a hand in - is that I find RT=_____. I have a series parallel combination circut that starts in series with 2 parallel circuts R1 thru R6 they range from 500 ohms to 3.5K ohm resistors across the circut. My IT=10ma. that's all the information I have been given. The formula I surmise is as follows.....R5 + R6 =_____ = 1 OVER = REQ1 1/ + R4 1/ = 1/ = _________ + R2 + R3 = REQ2 1/ + R1 1/ , 1/ = RT.

If someone could tell me if I am getting this right the RT I am getting is 656.03 -6 ohms....... is this possible with these values.... R1 = 3.5 K
R2 = 1.5 K
R3 = 1.2 K
R4 = 1 K
R5 = 500 OHM
R6 = 3.5 K
RT = _______

R1 and R4 are in parallel

Could some one please show me the correct formula? for step 1 of computing this circut.......I am just starting to learn this in IT school and I am having a hard time wrapping my head around this!

while this is a fundamental task in electronics the school I attend finds that for an IT student they need to have a basic electronics understanding of circut formulas and understanding..... Step two - figuring all the totals is challenging as well... for me.....I am by no means a brainiac and I am just learning so please be gentle in your responses....thank you in advance... for s/giggles the IT is 10ma if your so inclined......maybe easy for you but I having a hard time here......

thank you for all those equations and answers.....I was instructed by my instructor to start at the point furtherst from the power supply working backwards......to power supply........solving to one resistor.....r1 and r4 and r6 are the two in parallel and r2,r3,r5 are in series......r1 is next to power source before r2(top), r3 ( bottom) then r4 parallel with r5 on top and r6 in series.....hopefully this clarifys...all thus far were good, however 4 diffrent answers have given me 4 different ways to be just a bit more confused.......the first answer seem ed like it was the best....then I read the second....and scratched my head.......hearing what the instructor told me in the way to get RT.....the way I have figured it out is......RT=6.7K which would then show IT=10ma and then VT 67v....which makes my I1=19.14 ma....I don't see how I gained 9.14 ma going to I1 parallel circut.........so you can see.... I am stuck again........or maybe not seeing something right. HELPPPPPPPPPPPPPPPPP

Answer 1

Series:

R2 = 1500
R3 = 1200
R5 = 500
R6 = 3500

Total series R = 6700 ohms

Parallel:

R1 = 3500
R4 = 1000

1 / 4500 = 0.00022

6700 + 0.00022 = 6.70000022 k ohms

Answer 2

ok im not quite sure how the circuit is set up based on your description, but im gonna take a guess by the way you set up the formula that R2 and R3 are in series and are in parallel with R1. R5 and R6 are in series and are in parallel with R4. the R1||(R2+R3) resistor combination is then in series with R4||(R5+R6). so the formula for the circuit should start out like: Rt = (R1||(R2+R3)) + (R4||(R5+R6)).
hopefully this is how the circuit is set up.
so now to find Rt, start with finding the resistance of the R1||(R2+R3) resistor combination
R1=3500
R2=1500
R3=1200
R1||(R2+R3) = 1/((1/R1)+(1/(R2+R3))) = 1/((1/3500)+(1/(1500+1200))) = 1/((1/3500)+(1/2700)) = 1/((.0002857)+(.0003704) = 1/(.0006561) = 1524.16 ohms

now do the same for the R4||(R5+R6) resistor combination
R4=1000
R5=500
R6=3500
R4||(R5+R6) = 1/((1/R4)+(1/(R5+R6))) = 1/((1/1000)+(1/(500+3500))) = 1/((1/1000)+(1/4000)) = 1/((.001)+(.00025)) = 1/(.00125) = 800 ohms.

so now R1||(R2+R3) = 1524.16 ohms and R4||(R5+R6) = 800 ohms
Rt = R1||(R2+R3) + R4||(R5+R6)
Rt = 1524.16 + 800 = 2324.16 ohms.
Rt = 2324.16 ohms

hopefully this helps

Answer 3

For resistors in parallel, the equation is
1/Rt = 1/R1 + 1/R2 + R3.... + 1/Rn where
Rt = total resistance
R1, R2, R3 and Rn are the resistors in parallel.

When you have two resistors in parallel, you can make the solution can be simplified to product over the sum:

(R1 * R2) / (R1 + R2)

Resistors in series are just added together.

The way to solve these circuits is to keep simplifying them. For example, R1 is in parallel with R2 and that parallel pair is in series with R3, first calculate a new resistor to replace R1 and R2 in parallel. Then you have R(new) and R3 in series and you just add them.

If you have something like R1 and R2 in series and that series pair is in parallel with with a single resistor R3, first replace R1 and R2 with anequivalent resistor R(new). Now you have R(new) in parallel with R3 so using the product over the sum, calculate the parallel resistance.

Just draw out your circuit and keep simplifying until you have a single equivalent resistance.

Answer 4

sure. in accordance to the Kirchoff regulation, the summation of the present entering into and going out of a think approximately circuit which incorporate in user-friendly terms resistors even they are parallel or sequence is 0. a similar applies to circuits with complicated currents. complicated currents are currents passing via inductor and capacitor.

Answer 5

those answers were long, so short

adding series resistors: Rtotal = R1+R2+R3+...

adding parallel resistors: Rtotal = 1 / ((1/R1) + (1/R2) + (1/R3))