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- The current division equation (Equation
) does not include
the resistance of the ammeter. Let the internal resistance of the ammeter be
. Write the expression for the current through
,
, including
the resistance of the meter assuming that the ammeter is being used to measure
the current
. Then take the limit of this expression as the ammeter internal
resistance goes to zero, showing that the limit is given by Equation
.
- The voltage source and 10 k
resistor in Figure
form
an approximate current source for small load resistances.
If the voltage source and 10 k
resistor
formed an ideal current source, then the current
would be constant,
independent of the resistances of
and
, which is certainly
not the case. Consider the parallel
combination of
and
as a single resistance
. If
is
small compared to 10 k
, then the current
will be very nearly
1 mA (Recall that
) independent of
. Calculate the range
of values of
such that the current
will deviate from 1 mA by no more than
5%.
-
Consider the circuit shown in Figure
.
Suppose you want to know the value of all voltages and currents in the circuit.
Assume that you know nothing at all about the resistor values.
You want the results to be as accurate as possible. You have a multimeter
that you may use as either a voltmeter or an ammeter.
Explain the sequence of measurements that you make. Comment on your
level of confidence that your results are accurate. Don't forget that you
have Ohm's Law and Kirchhoff's Laws that may be used.
Figure: Resistive Network
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