1. Oct 10, 2005

### Salman

PROBLEM:the current at the terminals of a certain current source is measured with an ammeter having an internal resistance $$R_i=10 ohms$$ and is found to be $$11.988 mA$$ ;adding a $$1.2 kilo -ohms$$ resistance between the source terminals causes the ammeter reading to drop to $$11.889 mA$$.Find $$i_s$$ and $$R_s$$

NOTE:$$i_s$$ and $$R_s$$ constitute a practical current source.i.e $$R_s$$ is the internalresistance of the source $$i_s$$

note:my answers don't quite match with the answer at the back of the textbook....
my calulations: $$i_s=11.997 mA$$ and $$R_s=13200ohms$$
answers in the answer booklet:$$i_s=12mA$$ and $$R_s=10kilohms$$:uhh:

2. Oct 11, 2005

### CarlB

I get the two voltages at the top of the ammeter as .11988V and then .11889V. The two currents are then 11.988mA and 11.889mA+.11889V/1.2kOhms.

The change in voltage was .00099V
The change in current was .000075mA = .000000075A

So Rs = .00099V/.000000075A = 13,200 ohms just like you got. Maybe they're rounding to a single significant figure.

Then Is = 11.988 mA + .11988V / 13.2 Kohms = 11.99708 mA