# Finding Resistor Values

Discussion in 'Homework Help' started by WRBWRX04, May 20, 2014.

1. ### WRBWRX04 Thread Starter New Member

May 15, 2014
1
0
I am suppose to find the Resistor values of R1, R2, and R3. I'm not sure if Im on the right track but, When the switch is open, the LED is on.

Since the LED is on that will receive a Volts high value of 2.4Volts in addition to the whats left from the 5Volts above. is that correct?

The current of 30mA is when the switch is open, if the switch was then it would short out the led and no current would flow through that branch. is this correct?

I know there is a voltage drop of 1.7 volts across the led but I don't understand the Vhigh and Vlow values in the diagram is that listed wheat it is or is it in addition to the 5volts?

Can't get the picture to show up for some reason but I include a link from my drop box
https://www.dropbox.com/s/1isiil2w163k38o/ExtraCR.jpg

Last edited: May 20, 2014
2. ### crutschow Expert

Mar 14, 2008
13,509
3,385
I believe Vhigh is the node (wire) voltage with the switch open and Vlow is with the switch closed.

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,515
515
What is the voltage to turn led ON?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,515
515
When switch is open, you have voltage divider. The circuit itself is one loop:
From Voltage Source to R1, from R1 to LED, from LED to R2, from R2 to Voltage Source.

You say that the current in this case is 30 mA. Then you have:
R1(30 mA)+2.4 volts+R2(30 mA)=5 volts
Problem here is that you have two unknowns, R1 and R2, and only one equation, this is impossible to solve because you need at least one equation for each unknown. You need more equations. So we move to the next situation.

The switch is closed. Now you have a current divider and you also have two loops.
If the voltage across the led is given and it is 0.4 volts. I am not sure how to treat led here. Do I represent it with 0.4 volt Voltage Source? If yes, then just use Mesh Current Method or Node Voltage Method to setup more equations. At this point you have 3 unknowns (R1, R2, R3) and you need 3 equations. We already got 1 equation from earlier, now we need two more. Also, at some point you are going to have to account for the currents in the current divider. Since you have two loops, you have 1 current entering the current divider (30 mA) and two currents (i1 and i2) exiting. So your accounting for currents will be something like this: i1+i2=30 mA.

Thus far it looks like you have 5 unknowns (R1, R2, R3, i1, i2). Find them you need 5 equations. You have 2 equations:
i1+i2=30 mA
R1(30 mA)+2.4 volts+R2(30 mA)=5 volts
You need 3 more equations.

5. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
This equation is wrong. The 2.4V is pretty clearly the voltage on that node (I'll call it the output voltage for convenience) when the switch is open. It is NOT the voltage drop across the LED since that is stated as being 1.7V.

Just knowing that the output voltage is 2.4V when the current is 30mA is enough to size R1. Similarly, just knowing that the sum of the LED voltage and the voltage across R2 is 2.4V when 30mA is enough to size R2.

All that is left is R3. Since we have R1 it is an easy matter to find out how much current must flow to get the output voltage down to 0.4V. Since the LED will be off under that condition, we know that all of that current must flow through R3 while producing 0.4V.

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227

No that is not correct. Kirchoff's voltage law must be satisfied.