Calculating Current and Voltage in a Circuit with Multiple Resistors

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SUMMARY

The discussion focuses on calculating current and voltage in a circuit containing multiple resistors, specifically a 5Ω, 4Ω, 3Ω, 2Ω, and 1Ω resistor. Participants emphasize the importance of using Ohm's Law and current division principles to determine voltages across each resistor. The 4Ω and 5Ω resistors are confirmed to be in parallel, sharing the same voltage. Participants also highlight the necessity of accurately drawing circuit diagrams to simplify analysis and calculations.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of current division in parallel circuits
  • Familiarity with voltage division in series circuits
  • Ability to interpret and draw circuit diagrams accurately
NEXT STEPS
  • Learn how to apply Ohm's Law to calculate voltage drops across resistors.
  • Study the principles of current division for resistors in parallel.
  • Explore voltage division techniques for resistors in series.
  • Practice drawing and analyzing complex circuit diagrams for clarity.
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone looking to deepen their understanding of resistor networks and circuit analysis techniques.

  • #31
William White said:
you don't know what the voltage across the resistors are because you have not drawn it properly!

have you followed the step-by-step instructions I gave you?

ie have you looked at my drawing and drawn your own?
- unless you have the mental insight of a genius its essential to always - always - draw the problem clearly and neatly. Once this is done, you will find the problem virtually solves itself. The scribble you provided is creating you more problems than it is solving!
william at this point simply drawing it out is not going to solve the problem, as Op has your drawing to look at
 
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  • #32
nothing909 said:
no, i have followed the steps and redrew it but i still don't get it
check post 29
 
  • #33
donpacino said:
what is the relationship between the 5 ohm resistor and the four ohm resistor. that should give you some insiight into the relation between the four ohm resistor and the (1 & 2) ohm resistors.
the
donpacino said:
check post 29
the 4 ohm resistor is in parallel with the 5, yeah, so are all 3 of them in parallel with each other?
 
  • #34
nothing909 said:
the

the 4 ohm resistor is in parallel with the 5, yeah, so are all 3 of them in parallel with each other?
yup! if you have any doubts, look up the definition of parallel resistors. eventually it will be intuitive
 
  • #35
nothing909 said:
no, i have followed the steps and redrew it but i still don't get it

okay, so you know the total currentif you have drawn the circuit properly, you will see that the current is going through three parallel paths, top middle bottom

do you know the rules for dividing the current through the top middle and bottom paths?if the top path is 3ohms
the middle path is 4 ohms
and the bottom path is 5 ohms

and the total current is 14A

what current is going though each path? The lower the resistance the MORE current, so expect the largest amount to go through the top path, and the least in the bottom path.

when you know these currents, tell me what they are.
 
  • #36
okay, thank you, but i feel like an idiot still asking question when u guys have explained all this like 10 times.

if the voltage for the 4 ohm resistor is 17.09v, that means the voltage across the 5 resistor and the 1 ohm resistor are all 17.09, yeah? so what's the voltage across the 2 ohm resistor?
 
  • #37
William White said:
okay, so you know the total currentif you have drawn the circuit properly, you will see that the current is going through three parallel paths, top middle bottom

do you know the rules for dividing the current through the top middle and bottom paths?if the top path is 3ohms
the middle path is 4 ohms
and the bottom path is 5 ohms

and the total current is 14A

what current is going though each path? The lower the resistance the MORE current, so expect the largest amount to go through the top path, and the least in the bottom path.

when you know these currents, tell me what they are.
i don't understand how to work it out. do i need to just do something like voltage division but instead with current?
 
  • #38
nothing909 said:
okay, thank you, but i feel like an idiot still asking question when u guys have explained all this like 10 times.

if the voltage for the 4 ohm resistor is 17.09v, that means the voltage across the 5 resistor and the 1 ohm resistor are all 17.09, yeah? so what's the voltage across the 2 ohm resistor?
close but no... that 19.09 V is across the series combination of the 1 and 2 ohm resistors.

and again don't feel bad for asking questions
 
  • #39
so 17.09 is the Vin? so i just do 17.09 x 1/3 and 17.09 x 2/3 and that'll give me the voltages for both of the individual resistors?
 
  • #40
nothing909 said:
i don't understand how to work it out. do i need to just do something like voltage division but instead with current?

so, this is the key to solving the problem;

the current through the resistor is inversly proportional: less resistance more current.

you have 14A

If you have a 2+1 = 3 ohm resistance in the top path. 4 ohm resistor in the middle
5 ohm in the bottom

can you show that the currents are
6A
4.5A
3.5A

?
 
  • #41
William White said:
so, this is the key to solving the problem;

the current through the resistor is inversly proportional: less resistance more current.

you have 14A

If you have a 2+1 = 3 ohm resistance in the top path. 4 ohm resistor in the middle
5 ohm in the bottom

can you show that the currents are
6A
4.5A
3.5A

?
look at my above post. is that right what i said?
 
  • #42
yes, the voltage across the three parallel paths is 17.9 V...

so you know the resistances...so what are the currents through each path?
 
  • #43
William White said:
yes, the voltage across the three parallel paths is 17.9 V...

so you know the resistances...so what are the currents through each path?
The currents are just I=V/R after I've worked out the voltages, yeah?
 
  • #44
nothing909 said:
so 17.09 is the Vin? so i just do 17.09 x 1/3 and 17.09 x 2/3 and that'll give me the voltages for both of the individual resistors?
exactly
 
  • #45
Yes,

and you have told me the voltage drop across the three paths - it is 17.9 V

so now tell me the current through each resistor...
 
  • #46
the current across the 4 ohm resistor is 71.4
the current across the 5 ohm resistor is 89.55
the current across the 1 ohm resistor is 5.97
the current across the 2 ohm resistor is 5.97
 
  • #47
you have already told us the correct TOTAL current, which is 14A

all your currents must add up to this! you have multiplied rather than divided! i = v / R

you have 17.9V across a 5 ohm resistor...that is 3.58 A

what are the others?
 
  • #48
wait, i don't understand. is my answer for the 1 and 2 ohm resistors correct?

i did for the 1 ohm resistor: 17.91x1/3, then I=5.97/1

for the 2 ohm resistor : 17.91x2/3, then I=11.94/2
 
  • #49
yes;

The 1 and 2 ohm resistors are in series - the current through them is the same.

therefore they are equivalent to a 3 ohm resistor

17.9 V / 3 ohm = 6A
 
  • #50
yeah sorry

for the 4 ohm it's 4.48
for the 5 ohm it's 3.58
 
  • #51
4.48 + 3.58 + 5.97 = 14.03
 
  • #52
looks good man...
my recommendation, work a few more practice problems like this!
 
  • #53
k, wait, just one more question before i go. how did will work out the total current was 14A?
 
  • #54
nothing909 said:
4.48 + 3.58 + 5.97 = 14.03

yes!

now, you can see the key is to have a good drawing!

10 minutes on a good drawing saves you hours of work!
 
  • #55
nothing909 said:
k, wait, just one more question before i go. how did will work out the total current was 14A?

you work out the total resistance of the circuit

look at my drawing

you have

3ohms (the resistor in series) +
1 / ( 1/(1+2) + 1/4 + 1/5 ) = 1.3 ohms (this is the rule for resistors in parallel ... you remember learn this one!)

total R = 3 + 1.3 = 4.3 ohms

total current = voltage / total R
= 60 / 4.3
= 14A
 
  • #56
omg, that was such a silly question, i worked that out myself right at the very start...

anyway thanks for all the help, that simple question was giving me a lot of stress.
 

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