Question

Hey looking for some help on a problem that I cannot seem to figure out. Medal for answer and explanation of course. And the topic is of circuits and electcity, will post details below. Thanks!

Answer 1

The current passing through R1 does not change regardless of if Switch 1 and switch 2 are both open, or both switches are closed. Looking to find Rx

Answer 2

So far I have tried a few different things but nothing that leads to the answer. When both switches are open, the total R=298ohm, when both are closed it is 106+ (1/Rx+1/126)^-1

Answer 3

What I gather is that when switches are open we have to go through R3 but also R1 is in series, when closed, we can bypass R3 but R1 is now in parrallel with Rx.

Answer 4

That is correct. When both switches are open, the current through R1 = 10 / 298.

When both switches are closed, R3 gets shorted out. The effective resistance is:
106+ (1/Rx+1/126)^-1
Find the total current drawn and the current passing through R1.

Equate the current in the two cases and solve for Rx.

Answer 5

would the current drawn be 10v-i1(106ohm)-i2(126ohm)-i3(66ohm)? from when the switches are both open?

Answer 6

If you find the current using 298 and V= 10 you solve i=0.0356 Amperes, is this the value I need to work with? Do I use this i to calculate voltage drops?

Answer 7

In which case, 0.0356xR2=3.8v, so 6.2 V leftover, dividing 6.2v by 126ohm says i=0.0492 A?

Answer 8

When the switches are open, R2, R1 and R3 are in series and the same current flows through all three of them No need to calculate voltage drops.
When the switches are closed, you need to find the total current drawn in terms of Rx. Then find the voltage drop across R2. 10V - the voltage drop across R2 will be the voltage drop across R1 || Rx (that is, R1 in parallel to Rx). From this voltage drop you can calculate the current through R1. Voltage drop / R1. Equate the currents in the two cases and solve for Rx.

Answer 9

Is 205 ohms one of the answer choices?

Answer 10

It isnt multiple choice,

Answer 11

I did not double-check my calculations but that is what I got for Rx.

Answer 12

Okay, Im wondering the method to find the total current drawn by rx?

Answer 13

Yea i tried 205, it wasnt that

Answer 14

No need to find current through Rx.
Let i be the total current drawn. Then i * 106 will be the voltage drop across R2.
10 - 106i will be the voltage drop across the parallel circuit R1 || Rx.
Current through R1 = Voltage drop across R1 / Resistance = (10 - 106 * i) / 126

Answer 15

and are you getting the value of 0.035 as i total?

Answer 16

Did not solve for i. Kept them all in terms of Rx and just solved for Rx.

Answer 17

Hm do you have an idea of why 205 did not work?

Answer 18

There were a lot of big numbers and I did rapid fast calculations and that is why I left open the possibility of a calculation mistake.

Answer 19

Oh okay no problems. What is the equation you can use to put in terms of Rx?

Answer 20

\[10 - \frac{ 1060(126+x) }{ 13356 + 232x } = \frac{ 1260 }{ 298 }\]

Answer 21

okay and that is the voltage drop correct?

Answer 22

Starting from the beginning:
Effective resistance when both switches are closed:
106 + 126x/(126+x) = (13356 + 106x + 126x) / (126 + x) =
(232x + 13356) / (126 + x)
Total current = V/R = 10 * (126+x) / (232x + 13356)

Answer 23

Sorry im not even to clear where the 13356 is coming from?

Answer 24

making (126+x) the common denominator. Sp 106 * (126+x)

Answer 25

I see, so the 106 + 126x/(126+x) is from when both switches are closed?

Answer 26

which makes sense because it ignores the 66

Answer 27

Yes. If A and B are resistors in parallel, then 1/R = 1/A + 1/B = (A+B)/AB
So R = AB/(A+B)

So effective resistance = R2 + R1*x/(R1+x) = 106 + 126x/(126+x) =
(232x + 13356) / (126 + x) (after simplifying)

Answer 28

Okay great so I understand that part! Now just wondering how you use this to proceed

Answer 29

Total current drawn by the circuit = V/R = 10 * (126+x) / (232x + 13356)
Voltage drop across R2 = current * R2 = 106 * 10 * (126+x) / (232x + 13356)
The remaining voltage = battery voltage - voltage drop across R2 =
10 - 1060 * (126+x) / (232x + 13356)
Simplify above.
This will be the voltage drop across R1 (and Rx too).
Current through R1 = voltage drop across R1 * R1

Equate current through R1 in both cases and solve for x.

Answer 30

I am getting Rx = 202.36 ohms.
It may have to be rounded down to 202 ohms depending on what degree of accuracy they want.

Answer 31

I see , so what do you do with the term 10 - 1060 * (126+x) / (232x + 13356) . Just multiply it by 126? (to get the current through R1)

Answer 32

So in that cause when you solve for remaining voltage, you get 4.23 V, if you times that by 126ohm you get 532.8 A? Is that correct @ranga

Answer 33

Sorry i keep asking just trying to work through it all :D