# Voltage Divider Design

Discussion in 'Homework Help' started by yan500, Jul 17, 2011.

1. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hi all,

Hope everyone's having a great weekend. I have a problem that I'm trying to solve and it's tripping me up. http://www.allaboutcircuits.com/worksheets/e_divide.html

It is Question 35 on the worksheet located at the link above. Since the problem states that you can assume there is a bleed current of 1.5 mA, I calculated the total current in the circuit to be 13.5 mA. I believe that this is what's throwing me off. Can someone tell me what the total current in the circuit is and how they calculated it?

Thanks!

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
To get I_total you simply need to add the load current plus the bleeding current.

3. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Right, but for R3 + Load 1 you get 4.5 mA, which gives you the right answer for R3. But shouldn't the total of R2 + Load 2 give you 4.5 mA as well because of the loop rule?

4. ### ian_gregg New Member

Jan 19, 2011
21
0
There is 1.5mA through R3, 4.5mA through R2 and 16.5mA through R1, you know the voltages, V=IR and away you go.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
I don't understand you problem
R2 = (9V - 4.8V)/4.5mA = 933.3

6. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Ah, I see. Thank you. But how did you know to subtract 9 from 4.8 to get the voltage going through R2?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Because I know Ohms law and KVL/KCL all too well.
So the voltage drop across R2 must be equal 9V - 4.8V

Feb 17, 2009
3,990
1,115