Question

Can anybody please help me with this?

Answer 1

well, check your Ohm's law and formula of resistance first :)

Answer 2

V=I R

Answer 3

is easier to read if it were right side up btw

Answer 4

resistance in series
\[R_{eq}=R_1+R_2+...\]
resistance in parallel
\[\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} + ...\]

Answer 5

How to find Rc? can you explain?

Answer 6

by applying ohms law

you know V and you know I

and R = Rc

Answer 7

V - voltage from battery
I- current read from ammeter

Answer 8

and how do I apply it? can you please explain:(?

Answer 9

i cannot explain it any further than what i've already said just substitute the values and solve for Rc...

Answer 10

ohms law is just a simple equation
V=I R

Answer 11

6=1 x R ? is this correct @completeidiot ?

Answer 12

yes...

Answer 13

so R is 6 right?

Answer 14

V = I R, 6 = 1 * R, yes, total R must be 6

Answer 15

if you have R1=50 OHms and you put R2=1 Ohms in parallel,
the total resistance is about 1 Ohms only!(even some less than that)

Answer 16

And how do I find X?

Answer 17

so you need that all the resistances combined are =6

what they want you to do is to make a replacement resistor instead of those three and their configuration.

the one total resistor (in between battery and monitoring) should be 6

Answer 18

ok so there's one in series that adds 3 Ohms and you can't do anything about that one. the resistance is already 3 Ohms at least because that has been fixed in the image.

The question is, how can we make 3 + two parallel = 6

Answer 19

ok so far?

Answer 20

all the branched, layered resistors do have the effect of ONE SIMPLE resistor

Answer 21

I have no idea :( can you explain?

Answer 22

yes

Answer 23

the voltage is considered to be stable no matter how much we draw

Answer 24

ok so if we have one resistor the voltage of 6V is able to push through it a certain amount

Answer 25

if we put two resistors in parallel --- we can push the double amount! since there is now two congested pathways, not one.

Answer 26

this is resistors in parallel, every additional parallel resistor decreases resistance

Answer 27

because a resistor can be seen from two perspectives; how much does it block, how much does it let through. in parallel, the latter is important.

Answer 28

Law: resistors in series do add.

Answer 29

we have 3 Ohm + the parallel branch

Answer 30

...so the parallel branch must be 3 Ohm, also, to get a total 3+3=6 Ohm

Answer 31

the 3 Ohm and the parallel branch are in series, right?

Answer 32

yes

Answer 33

yes and the total must be 6 Ohms, because of V = R I and 6 = R * 1

Answer 34

so parallel branch needs to be 3 Ohms

Answer 35

yes!

Answer 36

OK, the rule for parallel resistors is actually the following thing:
\[R_{total} = \frac{ R _{1}\times R_{2}}{ R_{1}+R_{2} }\]

Answer 37

we need
\[3 = \frac{ 12 \times R_{2}}{ 12+R_{2} }\] , R2 is X

Answer 38

3=2.4 is what I got

Answer 39

hmm

Answer 40

\[3 = \frac{ 12 \times R2}{ 12 } + \frac{ 12 \times R2}{ R2 } \]

Answer 41

\[3 = \frac{ 12 \times R2}{ 12 } + 12\] i just cancelled the R2 here

Answer 42

no

Answer 43

so x would be?

Answer 44

we have to solve that equation

Answer 45

the one

Answer 46

\[3 = \frac{ 12 \times X}{ 12 + X }\]
fluttering algebra

Answer 47

how to solvew for x??

Answer 48

help

Answer 49

I am not sure :S

Answer 50

me neither,ill create a question......

Answer 51

ok,thanks:)

Answer 52

\[3 = \frac{ 12 }{ 12 + x } \times \frac{ x }{ 1 }\]

\[3 (12+x) = 12 \times \frac{ x }{ 1 }\]

\[36+3x = 12 x
\]
36 = 9 x

x=4