DirectCurrent Circuits
2613
(b)
For a 500mV voltmeter, the resistances are in series and the current is the same through each:
( )
cs
ab
VI
R
R
=
+
and
3
sc
6
500 10
V
25.0
975
.
500 10
A
ab
V
RR
I
−
−
×
=−
=
−
Ω
=Ω
×
EVALUATE:
The equivalent resistance of the voltmeter is
eq
s
c
1000
R
=
+=
Ω
. The equivalent resistance of
the ammeter is given by
eq
sh
c
111
R
=+
and
eq
0.625
R
=
Ω
. The voltmeter is a highresistance device and the
ammeter is a lowresistance device.
26.32.
IDENTIFY:
The galvanometer is represented in the circuit as a resistance
R
c
. Use the junction rule to relate the
current through the galvanometer and the current through the shunt resistor. The voltage drop across each parallel
path is the same; use this to write an equation for the resistance
R
.
SET UP:
The circuit is sketched in Figure 26.32.
Figure 26.32
We want that
a
20.0 A
I
=
in the external circuit to produce
fs
0.0224 A
I
=
through the galvanometer coil.
EXECUTE:
Applying the junction rule to point
a
gives
af
ss
h
0
III
−
−=
sh
a
fs
20.0 A
0.0224 A
19.98 A
II
I
=−=
−
=
The potential difference
ab
V
between points
a
and
b
must be the same for both paths between these two points:
()
fs
c
sh
sh
IRR IR
(
)
sh
sh
c
fs
19.98 A 0.0250
9.36
22.30
9.36
12.9
0.0224 A
IR
I
Ω
=
−
Ω
−
Ω
=
Ω
EVALUATE:
sh
c
;
R
<<
+
most of the current goes through the shunt. Adding
R
decreases the fraction of the
current that goes through
c
.
R
26.33.
IDENTIFY:
The meter introduces resistance into the circuit, which affects the current through the 5.00k
Ω
resistor
and hence the potential drop across it.
SET UP:
Use Ohm’s law to find the current through the 5.00k
Ω
resistor and then the potential drop across it.
EXECUTE:
(a)
The parallel resistance with the voltmeter is 3.33 k
Ω
, so the total equivalent resistance across the
battery is 9.33 k
Ω
, giving
I
= (50.0 V)/(9.33 k
Ω
) = 5.36 mA. Ohm’s law gives the potential drop across the
5.00k
Ω
resistor:
V
5 k
Ω
= (3.33 k
Ω
)(5.36 mA) = 17.9 V
(b)
The current in the circuit is now
I
= (50.0 V)/(11.0 k
Ω
) = 4.55 mA.
V
5 k
Ω
= (5.00 k
Ω
)(4.55 mA) = 22.7 V.
(c)
% error = (22.7 V – 17.9 V)/(22.7 V) = 0.214 = 21.4%. (We carried extra decimal places for accuracy since we
had to subtract our answers.)
EVALUATE:
The presence of the meter made a very large percent error in the reading of the “true” potential
across the resistor.
26.34.
IDENTIFY:
The resistance of the galvanometer can alter the resistance in a circuit.
SET UP:
The shunt is in parallel with the galvanometer, so we find the parallel resistance of the ammeter. Then
use Ohm’s law to find the current in the circuit.
EXECUTE:
(a)
The resistance of the ammeter is given by 1/
R
A
= 1/(1.00
Ω
) + 1/(25.0
Ω
), so
R
A
= 0.962
Ω
. The
current through the ammeter, and hence the current it measures, is
I
=
V
/
R
= (25.0 V)/(15.96
Ω
) = 1.57 A.
(b)
Now there is no meter in the circuit, so the total resistance is only 15.0
Ω
.
I
= (25.0 V)/(15.0
Ω
) = 1.67 A
(c)
(1.67 A – 1.57 A)/(1.67 A) = 0.060 = 6.0%
EVALUATE:
A 1
Ω
shunt can introduce noticeable error in the measurement of an ammeter.
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 Spring '06
 Buchler
 Physics, Exponential Function, Current, Resistance, Resistor, Electrical resistance, Electrical impedance, Series and parallel circuits

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