Engineering Ohm's Law - Query regarding Current flowing in Circuit

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SUMMARY

This discussion focuses on applying Ohm's Law to determine equivalent resistance and current flow in a circuit. Participants utilize specific resistor values, such as RA = 3 kΩ and R_c = 5.5 kΩ, to calculate currents through various components. The calculations involve determining voltage drops and current splits, with a key formula being I = V/R. The conversation emphasizes the importance of clear notation and accurate unit representation in circuit analysis.

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hockeynut
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Homework Statement


(i) Using Ohm’s Law, determine the equivalent resistance across the battery (I think I've worked this out but am not sure I'm right)

(ii) Determine all the currents flowing in the circuit (I'm unsure how to get his?)

Homework Equations


Ohms Law


The Attempt at a Solution


See attached jpg potential solution to (i)
 

Attachments

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Hi hockeynut! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Once you have determined RD, and you know RA, you can calculate the current drawn from the battery. It all goes through RA, but divides and is shared between R2 and R3, for example.
 
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Hi NascentOxygen,
Thank you for your advice. I will try what you have suggested.
Most appreciated.
 
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Thank you for the help with the previous posting. I have gotten to the point attached, but am unsure regarding the voltage (current) across the r3, r4, r5 , and r6. I think r1 and r2 are ok. Any help/advice would be greatly appreciated.
 

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You are going to encounter problems when you write equations such as:
I1 = 1.044 x 6 = 6.264V

If I is current, then it can't have units of volts. Though it turns out that this isn't a calculation for current, 1.044 x 6 is a determination of voltage. So the subject of the formula won't be I1.

Don't be afraid to write words in among your calculations. I always say the layout of your working should read like a well-written essay! To wit,

Using RA and RD I determined the current in RA to be 1.044A.

So with this current through RA voltage drop across RA = 6k x 1.044A = 6.264v

Leaving voltage across R2 = 9.0 - 6.264 = ...

With this voltage across R4, the current to ground via R4 = ...
leaving a current through R3 of ...

and so on. See how you go now. :wink:
 
Hi NascentOxygen, Thank you kindly. I will review it tonight and see how I go. :)
 
Hi Nascent Oxygen,
Am I correct then in saying that the voltage going through loop two is 6.264v and divinding that by 8.62 resistance gives me the voltage for these two resistors R3 & R4 and follow the same rule for the next 2k / 1k resistor to get the right answer?
 
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:confused:Hi,
Does this look anyway right?
:cry:
 

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The current through R1 is 1.044mA. This current causes a voltage drop across R1 leaving just 6.264v at the junction of R1-R2-R3. At this node, the current splits and 6.264/5 mA flows through R2 to ground.

So you shouldn't show all of that 1.044mA going into R3, some of it is lost to ground via R2.
 
  • #10
Hi Nascent
I went down the following route, see below workings:
I = 9v/8.62k = 1.044 x 〖10〗^(-3)

From the previous section,
R_c=5.5 kΩ
⇒I_1=(5.5/(5.5+5)) I_tot
⇒I_1=(5.5/(5.5+5))×1.044 mA
⇒I_1=0.546 mA
⇒I_2=I_tot-I_1
⇒I_2=1.044-0.498
⇒I_2=0.0498 mA

R_A=3 kΩ
⇒I_3=I_4=(3/(3+3)) I_1
⇒I_3=I_4=(1/2)×0.546
⇒I_3=I_4=0.273 mA

Not sure I'm right...
 
  • #11
hockeynut said:
I went down the following route, see below workings:
I = 9v/8.62k = 1.044 x 〖10〗^(-3)

From the previous section,
R_c=5.5 kΩ
⇒I_1=(5.5/(5.5+5)) I_tot
⇒I_1=(5.5/(5.5+5))×1.044 mA
⇒I_1=0.546 mA
(5.5/(5.5+5))×1.044 mA is the current to ground via R2.
⇒I_2=I_tot-I_1
⇒I_2=1.044-0.498
⇒I_2=0.0498 mA
No sure what this is.
R_A=3 kΩ
⇒I_3=I_4=(3/(3+3)) I_1
⇒I_3=I_4=(1/2)×0.546
⇒I_3=I_4=0.273 mA

Not sure I'm right...
Does the sum of the currents through R2, R4 and R6 equal 1.044 mA?
 
Last edited:

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