Question

In the circuit shown, a current of 0.33 A will pass through which of the following resistors?

R3
R1
R2
R4

Answer 1

R1?

Answer 2

@justjm if you could help as well?

Answer 3

I've been looking at this far too long and I'm stumped
Are you sure that the question/image are correct?

I'm not able to get 0.33 at all mhmm

Answer 4

Or if you could show how you got to that and I'm just completely overthinking it

Answer 5

actually nvm that r1 would be for 6.6 V because (0.78)(8.6) = 6.6

Answer 6

im not sure 0.33 :/

Answer 7

Are you sure you're not misreading it and we're supposed to find 0.83 and not 0.33

Answer 8

R1 and R4 and the parallel combo R2 R3 all have the same current through them (I did not calculate it......it will likely be r2 or r3 since you cannot choose multiple )

Answer 9

right?

Answer 10

idk

Answer 11

and no it is 0.33, its not an error :/

Answer 12

Yes you're right

So for parallel, the voltage is the same and the resistance would be
1/(1/13 + 1/18) = 7.55 ohms

And voltage is 15 V (stays same in parallel, changes in series)

So 7.55 + 8.5 + 3.2 = 19.25 ohms
V = IR
So current for the resistors in series will all be the same which would be 15/19.25 = 0.779 A

So it can't be resistor 1 or 4

Answer 13

I think we're not considering the voltage drop after resistor 1

Answer 14

hmm , okay yes that does make some more sense.. im honestly not sure

Answer 15

@imqwerty whenever you get a chance and you're free, if you could take a look at this. We could use your intelligence:))

Answer 16

would it be okay if i closed this question, just so i can upload more? until @imqwerty
esponds

Answer 17

responds

Answer 18

Yea, for sure

Answer 19

Circuits and all are always a little tricky >,<

Answer 20

very true! ive been having a lot of difficulty

Answer 21

\(\color{#0cbb34}{\text{Originally Posted by}}\) @TheSmartOne
Yes you're right

So for parallel, the voltage is the same and the resistance would be
1/(1/13 + 1/18) = 7.55 ohms

And voltage is 15 V (stays same in parallel, changes in series)

So 7.55 + 8.5 + 3.2 = 19.25 ohms
V = IR
So current for the resistors in series will all be the same which would be 15/19.25 = 0.779 A

So it can't be resistor 1 or 4
\(\color{#0cbb34}{\text{End of Quote}}\)
this is correct, it can't the resistor 1 or 4, so we need to check the current in resistors 3 and 2

Answer 22

a total current of 0.779Amps flows in the circuit

Answer 23

When the current arrives at the junction connecting wire of resistors 3 and 2, it splits up.
So a part of the 0.779 A goes into resistor 2 and the remaining goes into resistor 3

Answer 24

how to calculate the amount of current that goes into each resistor?
ez- the more the resistance the lesser the current.

current in r2 = \(\large {I \times \frac{r_2}{r_2+r_3}}\)

current in r3 = \(\large {I \times \frac{r_3}{r_2+r_3}}\)

where I is the total current arriving at the junction = 0.799 A

Answer 25

how did I get that formula to find the currents in each branch?
- the potential drop across r2 and r3 will be the same because they have the same endpoints
assume that the potential drop is \(V\)

so \(V = r_2 \times i_2\)
and \(V = r_3 \times i_3\)
also \(i_2 + i_3 = I \)

where \(i_2\) = current in resistor r2
and \(i_3 \) = current in resistor r3

solving the above equations you can get the expression for \(i_2 \) and \(i_3\) that i wrote in my previous message

Answer 26

you're a genius qwerts <3

Answer 27

master <3