Determine currents through all resistors

[boo_lufc]

1) Using Ohm's Law determine the eqv resistance across the battery.

2) Determine the currents through all resistors

 

[gneill]

Some people may not trust downloading .doc files from "strange" machines. Can you post your file as an image of some form?

 

[boo_lufc]

Hopefully this is better

 

[gneill]

Yup. Much better.

So, what have you tried so far? It looks as though you're going to have to analyze the circuit to find all the currents; what methods have you learned?

 

[boo_lufc]

I have tried to get the equivalent resistance by working out series values then parallel values working from right to left if that makes sense. I don't think there is any more to this part than this.
I then end up with one Reqv as 1.625k.
I assume I then use ohm's law to work out the current leaving the battery?
I have only just started learning the loop method for the other current values but am getting lost because of the 3 voltage sources as I haven't seen this before.

 

[boo_lufc]

ignore the three voltage source bit, I am looking at another problem aswel and sent you the wrong response

 

[gneill]

The posted circuit only appears to have one voltage source: a single 9V battery.

{EDIT: Ah. Just saw your last post. Okay, no three sources.}

Your value for the equivalent resistance looks fine, and yes, you can determine the current supplied by the battery using Ohm's law with the battery voltage and equivalent resistance.

Note that if you solve for the loop currents in order to find all the resistor currents you will have another method to determine the equivalent resistance using Ohm's law, since you will have the battery voltage and the current.

What equations have you written for the loops?

 

[boo_lufc]

Good point regarding "if you solve for the loop currents in order to find all the resistor currents you will have another method to determine the equivalent resistance using Ohm's law, since you will have the battery voltage and the current".
Will be good to be able to check.
I only have basic knowledge of the loop method and would I be right in saying the first loop would have equation: 9 - I1*R1 - I2*R2 = 0?
I then don't know what I would do in loop 2 regarding a supply voltage or would it just be something like: I2*R3 + I2*R4 + (I2-I1)R2 = 0 ?

 

[gneill]

The idea is to "walk" around the loop and tabulate the voltage drops (or rises) that occur due to any currents flowing the components.

So for the first loop, beginning with the battery's rise,

9V - I1*R1 - (I1 - I2)*R2 = 0

Note that R2 has two currents running through it: one from the first loop and one from the second.

 

[boo_lufc]

That now seems to make sense. When you sub in the R values of 1 would you get:
9 - 2I1 - I2 = 0 ?

I am still unsure about the other loops, would they be:
Loop 2 : (I1 - I2)*R2 - I2*R3 - (I2 - I3)*R4 = 0

Loop 3: (I2 - I3)*R4 - I3*R5 - I3*R6 = 0 ?

 

[boo_lufc]

Sorry for loop 1 I meant:
9 - 2I1 + I2 = 0

 

[gneill]

It looks like your equations are good. Can you solve them for I1, I2, and I3?

 

[boo_lufc]

for loop 3 I got I2 = 3*I3 then subbed this value into the eqns for loops 1 and 2.
I got I1 = 3.789, and negatives for I2 and I3
I2 = -1.421
I3 = -0.474

 

[gneill]

I think you'll want to check your algebra. All the currents should be positive (with the assumption of clockwise ==> positive). The initial equations look okay, so something unfortunate must have happened while you were solving them.

 

[boo_lufc]

I have tried it again and have
I1 = 5.5 (The same answer i achieve by v/reqv)
I2 = 2.077
I3 = 0.692

 

[gneill]

That looks better. Note that the resistances were all given in KΩ (thousands of Ohms), so your currents should be divided by 1000, or if you prefer, interpreted as milliamps (mA).

 

[boo_lufc]

yea i got the mA bit.
Would i be right in saying that to answer the question i.e. the current through all resistors using the values for I1, I2 and I3 I got would be:
R1 = I1 = 5.5mA
R2 = I1 - I2 = 5.5 - 2.077 = 3.423mA
R3 = I2 = 2.077mA
R4 = I2 - I3 = 2.077 - 0.692 = 1.385mA
R5 = I3 = 0.692mA
R6 = ?
I am unsure what happens with R5 and R6 as they are in series?

 

[gneill]

Looks good (but beware of being too free and easy with the equals sign; Resistances are not currents. Use a colon instead). Yes, R5 and R6 are in series, so they share the same current.

 

[boo_lufc]

ok thanks so because R5 and R6 are the same values by share do you mean they would both be 0.692 or that this value would be divided between the two and they would both be half this

 

[gneill]

It is the same current -- equal in all respects.

Components in series have the same current though them. Components in parallel have the same voltage across them.

 

[boo_lufc]

Ok thanks for all the guidance, that should be great now thanks