Numerical problem about circuits

[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

 

[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

Maybe your answers are right, except for the rounding.

Carl

 

[Salman]

ok thank u sir!