Numerical problem about circuits

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SUMMARY

The discussion centers on calculating the internal current source (i_s) and internal resistance (R_s) of a circuit using an ammeter with an internal resistance of 10 ohms. The initial current reading is 11.988 mA, which drops to 11.889 mA when a 1.2 kΩ resistor is added. The calculations provided yield i_s = 11.997 mA and R_s = 13,200 ohms, while the textbook answers are i_s = 12 mA and R_s = 10 kΩ. The discrepancy is attributed to rounding differences in the textbook.

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Electrical engineering students, circuit designers, and anyone involved in practical circuit analysis and measurement accuracy.

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

NOTE:[tex]i_s[/tex] and [tex]R_s[/tex] constitute a practical current source.i.e [tex]R_s[/tex] is the internalresistance of the source [tex]i_s[/tex]

note:my answers don't quite match with the answer at the back of the textbook...
my calulations: [tex]i_s=11.997 mA[/tex] and [tex]R_s=13200ohms[/tex]
answers in the answer booklet:[tex]i_s=12mA[/tex] and [tex]R_s=10kilohms[/tex]:rolleyes:
 
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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
 
ok thank u sir!:smile:
 

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