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Unformatted text preview: way, we see that
the potential rises when we cross the battery, and drops when we cross each resistor. The
sum of the potential rises must equal the sum of the potential drops, so we have that V = I1R1 + I R2
{ 14 2 3
24
Potential
rises (1) Potential
drops Solving Equation (1) for R2, we obtain
I 2 R2 = V − I1 R1 or R2 = V − I1 R1 75.0 V − ( 9.0 A ) ( 4.0 Ω )
=
= 6.5 Ω
I2
6 .0 A c. This time, we apply the loop rule to loop CAED (see the drawing), traversing it
clockwise. This procedure yields V = I 3 R3 + I 5 R5
{ 14243
Potential
rises (2) Potential
drops Solving Equation (2) for R3, we obtain
I 3 R3 = V − I 5 R5 87. or R3 = V − I 5 R5
I3 = 75.0 V − (15.0 A ) ( 2.2 Ω )
= 3.5 Ω
12.0 A R c = 50.0 Ω
SSM REASONING Since only 0.100 mA out of
G
the available 60.0 mA is needed to cause a fullscale
59.9 mA
deflection of the galvanometer, the shunt resistor must 0.100 mA
B
A
allow the excess current of 59.9 mA to detour around
Shunt
the meter coil, as the drawing at the right indicates.
resistor R
The value for the shunt resistance can be obtained by 60.0 mA
recognizing that the 50.0Ω coil resistance and the
shunt resistance are in parallel, both being connected
between points A and B in the drawing. Thus, the voltage across each resistance is the same. Chapter 20 Problems 1113 SOLUTION Expressing voltage as the product of current and resistance, we find that 0.100 × 10 A ) ( 50.0 Ω
)
(59.9 × 10 A ) ( R3 = (14444244443)
1444
2444
–3 –3 Voltage across shunt resistance Voltage across coil resistance ( 0.100 × 10 A ) (50.0 Ω ) =
R=
–3 0.0835 Ω
–3
59.9 × 10 A
______________________________________________________________________________
88. REASONING The following drawing shows a galvanometer that is being used as a
nondigital voltmeter to measure the potential difference between the points A and B. The
coil resistance is RC and the series resistance is R. The voltage VAB between the points A
and B is equal to the voltage across the coil resistance RC plus the voltage across the series
resistance R. R RC
G
I A B
VAB SOLUTION Ohm’s law (Equation 20.2) gives the voltage across the coil resistance as I RC
and that across the series resistor as I R. Thus, the voltage between A and B is
VAB = I RC + I R Solving this equation for R yields VAB − I RC 10.0 V − ( 0.400 × 10−3 A )( 60.0 Ω )
=
= 2.49 ×104 Ω
I
0.400 ×10−3 A
______________________________________________________________________________
R= 89. REASONING According to Ohm’s law and under fullscale conditions, the voltage V across
the equivalent resistance Req of the voltmeter is V = IReq, where I is the fullscale current of
the galvanometer. The fullscale current is not given in the problem statement. However, it
is the same for both voltmeters. By applying Ohm’s law to each voltmeter, we will be able to
eliminate the current algebraically and calculate the fullscale voltage of voltmeter B.
SOLUTION Applying Ohm’s law to each voltmeter under fullscale conditions, we obtain 1114 ELECTRIC CIRCUITS VA = IReq, A VB = IReq, B and Dividing these two equations and eliminating the fullscale current I give
VB
VA = IReq, B
IReq, A = Req, B
Req, A or VB = VA Req, B
Req, A = ( 50.0 V ) 1.44 × 105 Ω
2.40 × 105 Ω = 30.0 V 90. REASONING AND SOLUTION The voltages across the galvanometer and shunt
resistances are equal since they are in parallel, so
IgRg = IsRs Ig
Rs = I
s or Rg where Ig = 0.150 × 10–3 A and Is = 4.00 × 10–3 A – 0.150 × 10–3 A = 3.85 × 10–3 A. Then 0.150 × 10−3 A Rs = (12.0 Ω ) = 0.468 Ω 3.85 ×10−3 A The equivalent parallel resistance can be obtained as follows:
1
1
1
=
+
Rp 0.468 Ω 12.0 Ω or Rp = 0.450 Ω ______________________________________________________________________________
91. SSM REASONING AND SOLUTION For the 20.0 V scale V1 = I(R1 + Rc)
For the 30.0 V scale
V2 = I(R2 + Rc)
Subtracting and rearranging yields I= V2 − V1
30.0 V − 20.0 V
=
= 8.00 × 10 –3 A
R2 − R1 2930 Ω − 1680 Ω Substituting this value into either of the equations for V1 or V2 gives Rc = 820 Ω .
______________________________________________________________________________ Chapter 20 Problems 1115 92. REASONING AND SOLUTION
a. According to Ohm's law (Equation 20.2, V = IR ) the current in the circuit is I= V
V
=
R+ R 2R The voltage across either resistor is IR, so that we find
V 60.0 V
V IR = = 30.0 V
R = =
2
2 2R b. The voltmeter’s resistance is R v = V / I = (60.0 V)/(5.00 × 10 –3 A) = 12.0 × 10 3 Ω , and
this resistance is in parallel with the resistance R = 1550 Ω. The equivalent resistance of
this parallel combination can be obtained as follows: 1
11
=
+
Rp Rv R Rp = or RRv R + Rv = (1550 Ω ) (12.0 × 103 Ω )
1550 Ω + 12.0 × 103 Ω = 1370 Ω The voltage registered by the voltmeter is IRp, where I is the current supplied by the battery
to the series combination of the other 1550Ω resistor and Rp. According to Ohm’s law I= 60.0 V
= 0.0205 A
1550 Ω + 1370 Ω Thus, the voltage registered by the voltmeter is IR p = (0.0205 A )(1370 Ω ) = 28.1 V
______________________________________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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