# Linear Circuit - current through Load Resistor

1. Oct 24, 2010

### ponomous

1. The problem statement, all variables and given/known data

When a 2-kΩ load is connected to the terminals A–B of the network in the figure, the current through the load is IL = 10 mA. If a 10-kΩ load is connected to the terminals, IL = 6 mA. Find IL if a 20-kΩ load is connected to the terminals.

2. The attempt at a solution

I thought that I could use the slope between the two given currents and resistors ie (2k-10k)/(10 mA - 6 mA) and then substitute in the 20k resistor and the unknown current into the formula. However this doesnt give the correct answer of 4 mA (given in the book).

Could someone point me in the right direction?

2. Oct 24, 2010

### cepheid

Staff Emeritus
Welcome to PF!

The reason why your method didn't work is because the variation of load current with load resistance is NON-linear. The load current varies inversely with the load resistance, and the function I(R) ~ 1/R is decidedly non-linear.

Assuming that your network (which is not shown) consists of nothing more than an ideal voltage source with a series resistance (or, in other words, reducing it to its Thevenin equivalent), then the open circuit voltage between terminals A and B is Vth. Furthermore, if the series resistance is Rth, then when you place a load resistor RL across the output terminals, the current across the load will be given by:

IL(RL) = Vth/(RL + Rth)​

This is the function that determines how the load current varies with load resistance. Now, you have two unknowns: Vth and Rth. BUT, you also have two equations, since you know IL(2 kΩ) and IL(10 kΩ). Therefore, you can solve for these two unknowns, at which point you'll know everything there is to know about the circuit. From then on, it's just a matter of plugging RL = 20 kΩ into the function. I did it and got the right answer.

3. Oct 24, 2010

### ponomous

Hey cepheid,

I understand what you are saying, and it makes sense, but i have run through this now 8 times and I am always getting 2 mA instead of 4. I cant find anything in my work that is wrong, any ideas?

Thanks

4. Oct 24, 2010

### cepheid

Staff Emeritus
Yeah, show us your work. Post your calculations and the results you got for Vth and Rth. Maybe I can spot where you went wrong...

5. Oct 25, 2010

### ponomous

So when i run through the two formulas i end up with

20(2000) + 20 Rth = Vth
60(10,000) + 60Rth = Vth

Solve and get -14,000 = Rth

This gives me a Vth of -240,000

When i substitute i get IL(20,000) = -240,000/ (20,000 - 14,000)

Which gives 2 mA

6. Oct 25, 2010

### cepheid

Staff Emeritus
What are the numbers in red? Are they supposed to be currents? If so, what units are they in? The current in the first equation is supposed to be 10 mA, so that seems completely wrong. The current in the second equation is supposed to be 6 mA, so I cannot see how that becomes 60, unless if you are expressing things in units of 10-4 A

Also, here's a useful tip: you can just keep everything in the units in which they are given. This is possible because:

(1 V) / (1 kΩ) = 1 mA

(1 mA)(1 kΩ) = 1 V

7. Oct 25, 2010

### cepheid

Staff Emeritus
How?

What I mean is, even with the wrong numbers, I don't see how you get 2 mA.

Ohhh no! Wait, you realize that in my expression below

that the parentheses on the left hand side represent function notation and not multiplication, right??? So, the left hand side should be read as IL OF RL.

I'm sorry if I misled you. I just thought it would be obvious from the context, since the current, I, must have dimensions of V/R. Plus, I referred to it as a function several times.

EDIT: This blunder also explains why your numbers (that I quoted in red) are weird. They are IL*RL when in fact they should be just IL. Arrgggh....

Last edited: Oct 25, 2010
8. Oct 25, 2010

### ponomous

OHHHHH oops.... That is my fault for using notation that is nearly identical to the function notation.

Thanks, I have the right answer now.

I know on some forums they have a thank button or something, is there something like that here? You have been most helpful.

9. Oct 25, 2010

### cepheid

Staff Emeritus
Hey, there's no "thank you" button, but I'm glad to have been of help, and you're welcome.