
#1
Mar1412, 06:00 PM

P: 9

1. The problem statement, all variables and given/known data
(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?) 2. Relevant equations Ohms Law 3. The attempt at a solution See attached jpg potential solution to (i) 



#2
Mar1512, 10:31 AM

HW Helper
P: 4,716

Hi hockeynut!
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. 



#3
Mar1812, 07:16 AM

P: 9

Hi NascentOxygen,
Thank you for your advice. I will try what you have suggested. Most appreciated. 



#4
Mar1812, 11:26 AM

P: 9

Ohm's Law  Query regarding Current flowing in Circuit
Thank you for the help with the previous posting. I have gotten to the point attached, but am unsure regarding the voltage (current) accross the r3, r4, r5 , and r6. I think r1 and r2 are ok. Any help/advice would be greatly appreciated.




#5
Mar1812, 08:21 PM

HW Helper
P: 4,716

You are going to encounter problems when you write equations such as:
I_{1} = 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 I_{1}. Don't be afraid to write words in among your calculations. I always say the layout of your working should read like a wellwritten essay! To wit, Using R_{A} and R_{D} I determined the current in R_{A} to be 1.044A. So with this current through R_{A} voltage drop across R_{A} = 6k x 1.044A = 6.264v Leaving voltage across R_{2} = 9.0  6.264 = .... With this voltage across R_{4}, the current to ground via R_{4} = ..... leaving a current through R_{3} of .... and so on. See how you go now. 



#6
Mar1912, 08:58 AM

P: 9

Hi NascentOxygen, Thank you kindly. I will review it tonight and see how I go. :)




#7
Mar1912, 02:14 PM

P: 9

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? 



#8
Mar1912, 04:59 PM

P: 9

Hi,
Does this look anyway right? 



#9
Mar1912, 08:49 PM

HW Helper
P: 4,716

The current through R1 is 1.044mA. This current causes a voltage drop across R1 leaving just 6.264v at the junction of R1R2R3. 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
Mar2112, 08:05 AM

P: 9

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_totI_1 ⇒I_2=1.0440.498 ⇒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
Mar2112, 10:24 PM

HW Helper
P: 4,716




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