Download Chapter 28 - Direct Current Circuits

Chapter 28 - Direct Current Circuits
P28.1
(a)
(b)
P=
( ΔV ) 2
R
( 11.6 V )2
becomes
20.0 W =
so
R = 6.73 Ω
R
FIG. P28.1
ΔV = IR
so
11.6 V = I ( 6.73 Ω )
and
I = 1.72 A
ε = IR + Ir
so
15.0 V = 11.6 V + ( 1.72 A ) r
r = 1.97 Ω
P28.2
P28.3
The total resistance is R =
3.00 V
= 5.00 Ω .
0.600 A
(a)
Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω
(b)
Pbatteries ( 0.408 Ω ) I 2
=
= 0.081 6 = 8.16%
Ptotal
( 5.00 Ω ) I 2
(a)
Here ε = I ( R + r ) , so I =
ε
R+ r
=
FIG. P28.2
12.6 V
= 2.48 A .
5.00
Ω
+ 0.080 0 Ω )
(
Then, ΔV = IR = ( 2.48 A )( 5.00 Ω ) = 12.4 V .
(b)
Let I 1 and I 2 be the currents flowing through the battery and the
headlights, respectively.
Then,
I 1 = I 2 + 35.0 A , and ε − I 1r − I 2 r = 0
so
ε = ( I 2 + 35.0 A ) ( 0.080 0 Ω ) + I 2 ( 5.00 Ω ) = 12.6 V
giving
I 2 = 1.93 A
Thus,
ΔV2 = ( 1.93 A )( 5.00 Ω ) = 9.65 V
FIG. P28.3
P28.7
If we turn the given diagram on its side, we find that it is the same as
figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first
reduction is shown in (b). In addition, since the 10.0 Ω , 5.00 Ω , and
25.0 Ω resistors are then in parallel, we can solve for their equivalent
resistance as:
1
Req =
= 2.94 Ω
( 10.01 Ω + 5.001 Ω + 25.01 Ω )
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
ΔV
and
Next, we work backwards through the diagrams applying I =
R
ΔV = IR alternately to every resistor, real and equivalent. The 12.94 Ω
resistor is connected across 25.0 V, so the current through the battery in
every diagram is
ΔV
25.0 V
=
= 1.93 A
I=
R 12.94 Ω
In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to give
a potential difference of:
ΔV = IR = ( 1.93 A )( 2.94 Ω ) = 5.68 V
From figure (b), we see that this potential difference is the same across
ΔVab , the 10 Ω resistor, and the 5.00 Ω resistor.
Thus we have first found the answer to part (b), which is ΔVab = 5.68 V .
(a)
P28.10
FIG. P28.7
Using 2.00-Ω, 3.00-Ω, and 4.00-Ω resistors,
there are 7 series, 4 parallel, and 6 mixed
combinations:
Series
2.00 Ω
3.00 Ω
4.00 Ω
5.00 Ω
*P28.13
Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
ΔVab 5.68 V
I=
=
= 0.227 A = 227 mA
25.0 Ω
Rab
6.00 Ω
7.00 Ω
9.00 Ω
Parallel
0.923 Ω
1.20 Ω
1.33 Ω
1.71 Ω
Mixed
1.56 Ω
2.00 Ω
2.22 Ω
3.71 Ω
4.33 Ω
5.20 Ω
FIG. P28.10
The resistance between a and b decreases. Closing the switch opens a new path with
1
resistance of only 20 Ω. The original resistance is R +
= R + 50 Ω .
1
90 + 10
+
1
10 + 90
The final resistance is R +
1
1
1
+
90
1
+
10
1
+
10
We require R + 50 Ω = 2(R + 18 Ω)
1
= R + 18 Ω .
90
so R = 14.0 Ω
−1
P28.15
1 ⎞
⎛ 1
+
Rp = ⎜
⎟ = 0.750 Ω
3.00
1.00
⎝
⎠
Rs = ( 2.00 + 0.750 + 4.00) Ω = 6.75 Ω
I battery =
ΔV 18.0 V
=
= 2.67 A
Rs 6.75 Ω
P2 = ( 2.67 A ) ( 2.00 Ω )
P = I 2R :
2
P2 = 14.2 W in 2.00 Ω
P4 = ( 2.67 A ) ( 4.00 A ) = 28.4 W in 4.00 Ω
2
ΔV2 = ( 2.67 A )( 2.00 Ω ) = 5.33 V,
ΔV4 = ( 2.67 A )( 4.00 Ω ) = 10.67 V
ΔVp = 18.0 V − ΔV2 − ΔV4 = 2.00 V ( = ΔV3 = ΔV1 )
P3 =
P1 =
P28.16
( ΔV3 )
R3
2
=
( 2.00 V )2
3.00 Ω
( ΔV1 ) ( 2.00 V )2
R1
=
1.00 Ω
FIG. P28.15
= 1.33 W in 3.00 Ω
= 4.00 W in 1.00 Ω
+15.0 − ( 7.00) I 1 − ( 2.00)( 5.00) = 0
5.00 = 7.00I 1
so
I 1 = 0.714 A
so
I 2 = 1.29 A
I 3 = I 1 + I 2 = 2.00 A
0.714 + I 2 = 2.00
+ε − 2.00 ( 1.29) − 5.00 ( 2.00) = 0
*P28.20
(a)
ε
FIG. P28.16
= 12.6 V
The first equation represents
Kirchhoff’s loop theorem. We choose
to think of it as describing a
clockwise trip around the left-hand
loop in a circuit; see Figure (a). For
the right-hand loop see Figure (b).
The junctions must be between the
5.8 V and the 370 Ω and between the
370 Ω and the 150 Ω . Then we have
Figure (c). This is consistent with the
third equation,
I1
220 Ω
I2
370 Ω
5.8 V
Figure (a)
I3
I2
370 Ω
150 Ω
Figure (b)
3.10 V
I1 + I3 − I 2 = 0
5.8 V
I 2 = I1 + I 3
(b)
We substitute:
I1
220 Ω
I2
I3
150 Ω
370 Ω 3.10 V
−220I 1 + 5.8 − 370I 1 − 370I 3 = 0
+370I 1 + 370I 3 + 150I 3 − 3.1 = 0
Figure (c)
Next
FIG. P28.20
5.8 − 590 I1
I3 =
370
520
370 I1 +
( 5.8 − 590 I1 ) − 3.1 = 0
370
370 I1 + 8.15 − 829 I1 − 3.1 = 0
I1 =
I3 =
11.0 mA in the 220-Ω resistor and out of
5.05 V
=
the positive pole of the 5.8-V battery
459 Ω
5.8 − 590 ( 0.011 0 )
370
= −1.87 mA
The current is 1.87 mA in the 150-Ω resistor and out of the
negative pole of the 3.1-V battery.
I 2 = 11.0 − 1.87 = 9.13 mA in the 370-Ω resistor
*P28.21
Let I6 represent the current in the ammeter and the top 6-Ω resistor. The bottom 6-Ω
resistor has the same potential difference across it, so it carries an equal current. For
the top loop we have
6 V – 10 Ω I10 – 6 Ω I6 = 0
For the bottom loop, 4.5 – 5 I5 – 6 I6 = 0.
For the junctions on the left side, taken together, + I10 + I5 – I6 – I6 = 0.
We eliminate I10 = 0.6 – 0.6 I6 and
0.6 – 0.6 I6 + 0.9 – 1.2 I6 – 2 I6 = 0
I5 = 0.9 – 1.2 I6 by substitution:
I6 = 1.5/3.8 = 0.395 A
The loop theorem for the little loop containing the voltmeter gives
+ 6 V –ΔV – 4.5 V = 0
ΔV = 1.50 V
P28.22
Label the currents in the branches as shown in the first figure.
Reduce the circuit by combining the two parallel resistors as
shown in the second figure.
Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:
( 2.71R) I 1 + ( 1.71R) I 2 = 250
and
(1.71R) I 1 + ( 3.71R) I 2 = 500
With R = 1 000 Ω , simultaneous solution of these equations yields:
I 1 = 10.0 mA
(a)
and
I 2 = 130.0 mA
From Figure (b),
Vc − Va = ( I 1 + I 2 ) ( 1.71R) = 240 V
Thus, from Figure (a),
I4 =
Vc − Va
240 V
=
= 60.0 mA
4R
4 000 Ω
Finally, applying Kirchhoff’s point rule at point a in Figure (a)
gives:
I = I 4 − I 1 = 60.0 mA − 10.0 mA = +50.0 mA
(b)
FIG. P28.22
,
I = 50.0 mA from point a to point e
or
P28.24
Using Kirchhoff’s rules,
12.0 − ( 0.010 0) I 1 − ( 0.060 0) I 3 = 0
10.0 + ( 1.00) I 2 − ( 0.060 0) I 3 = 0
and
I1 = I 2 + I3
12.0 − ( 0.010 0) I 2 − ( 0.070 0) I 3 = 0
10.0 + ( 1.00) I 2 − ( 0.060 0) I 3 = 0
FIG. P28.24
Solving simultaneously,
I 2 = 0.283 A dow nw ard in the dead battery
and
P28.31
I 3 = 171 A dow nw ard in the starter
The currents are forward in the live battery and in the starter, relative to normal
starting operation. The current is backward in the dead battery, tending to charge it up.
(a)
Call the potential at the left junction VL and at the right VR . After a
“long” time, the capacitor is fully charged.
VL = 8.00 V because of voltage divider:
10.0 V
= 2.00 A
5.00 Ω
VL = 10.0 V − ( 2.00 A )( 1.00 Ω ) = 8.00 V
IL =
Likewise,
2.00 Ω
⎛
⎞
VR = ⎜
⎟ ( 10.0 V ) = 2.00 V
⎝ 2.00 Ω + 8.00 Ω ⎠
or
IR =
10.0 V
= 1.00 A
10.0 Ω
VR = ( 10.0 V ) − ( 8.00 Ω )( 1.00 A ) = 2.00 V
Therefore,
ΔV = VL − VR = 8.00 − 2.00 = 6.00 V
FIG. P28.31(a)
(b)
R=
Redraw the circuit
1
= 3.60 Ω
Ω
+
1
9.00
(
) (1 6.00 Ω )
RC = 3.60 × 10−6 s
P28.32
1
10
and
e− t RC =
so
t = RC ln 10 = 8.29 μ s
(a)
τ = RC = ( 1.50 × 105 Ω )( 10.0 × 10−6 F) = 1.50 s
(b)
τ = ( 1.00 × 105 Ω )(10.0 × 10−6 F) = 1.00 s
(c)
The battery carries current
10.0 V
= 200 μ A
50.0 × 103 Ω
The 100 k Ω carries current of magnitude
⎛ 10.0 V ⎞ − t 1.00 s
I = I 0 e− t RC = ⎜
⎟e
3
⎝ 100 × 10 Ω ⎠
So the switch carries downward current
P28.35
FIG. P28.31(b)
200 μ A + ( 100 μ A ) e− t 1.00 s
Series Resistor → Voltmeter
ΔV = IR :
25.0 = 1.50 × 10−3 ( Rs + 75.0)
Rs = 16.6 k Ω
Solving,
FIG. P28.35
*P28.40
The current in the battery is
15 V
10 Ω +
= 1.15 A.
1
1
5Ω
+ 1
8Ω
The voltage across 5 Ω is 15 V – 10 Ω 1.15 A = 3.53 V.
(a) The current in it is 3.53 V/5 Ω = 0.706 A.
(b) P = 3.53 V 0.706 A = 2.49 W
(c) Only the circuit in Figure P28.40c requires the use of Kirchhoff's rules for solution.
In the other circuits the 5-Ω and 8-Ω resistors are still in parallel with each other.
(d) The power is lowest in Figure P28.40c. The circuits in Figures P28.40b and P28.40d
have in effect 30-V batteries driving the current.
*P28.42
I=
ε
R+ r
Let x ≡
, so P = I 2 R =
ε2
P
ε 2R
( R + r )2
or
, then ( R + r ) = xR or
2
With r = 1.20 Ω , this becomes
⎛ε2 ⎞
⎟R
⎝P ⎠
2
(R + r ) = ⎜
R2 + ( 2r − x ) R − r 2 = 0
R2 + ( 2.40 − x ) R − 1.44 = 0
which has solutions of
R=
(a)
With
ε
− ( 2.40 − x ) ±
( 2.40 − x )2 − 5.76
2
= 9.20 V and P = 12.8 W , x = 6.61 :
R=
+4.21 ±
( 4.21)2 − 5.76
= 3.84 Ω or
0.375 Ω . Either external
2
resistance extracts the same power from the battery.
(b)
For
ε
= 9.20 V and P = 21.2 W , x ≡
R=
+1.59 ±
ε2
P
= 3.99
( 1.59)2 − 5.76 1.59 ± −3.22
=
2
2
The equation for the load resistance yields a complex number, so
there is no resistance that will extract 21.2 W from this battery. The
maximum power output occurs when R = r = 1.20 Ω , and that maximum is
Pmax =
*P28.45
ε2
4r
= 17.6 W .
The charging current is given by 14.7 V – 13.2 V – I 0.85 Ω = 0
I = 1.76 A
The energy delivered by the 14.7 V supply is ΔVIt = 14.7 V 1.76 A 1.80 h (3 600 s/h) =
168 000 J
The energy stored in the battery is
13.2 V 1.76 A 1.8 (3 600 s) = 151 000 J
The same energy is released by the emf of the battery: 13.2 V (I) 7.3 (3 600 s) = 151 000
J
so the discharge current is I = 0.435 A
The load resistor is given by 13.2 V – (0.435 A) R – (0.435 A) 0.85 Ω = 0
R = (12.8 V)/0.435 A = 29.5 Ω
The energy delivered to the load is ΔVIt = I2R t = (0.435 A)2 29.5 Ω (7.3) 3 600 s = 147
000 J
The efficiency is 147 000 J/168 000 J = 0.873
*P28.50
(a) When the capacitor is fully charged, no current exists in its branch. The current in
the left resistors is 5 V/83 Ω = 0.060 2 A. The current in the right resistors is 5 V/(2 Ω
+ R).
Relative to the negative side of the battery, the left capacitor plate is at voltage
80 Ω (0.060 2 A) = 4.82 V .
The right plate is at R (5 V)/(2 Ω + R).
The voltage across the capacitor is 4.82 V – R (5 V)/(2 Ω + R).
The charge on the capacitor is
Q = 3 μF [4.82 V – R (5 V)/(2 Ω + R)] = (28.9 Ω – 0.542 R) μC/(2 Ω + R)
(b) With R = 10 Ω, Q = (28.9 –5.42) μC/(2 + 10) = 1.96 μC
(c) Yes. Q = 0 when 28.9 Ω – 0.542 R = 0
R = 53.3 Ω
(d) The maximum charge occurs for R = 0. It is 28.9/2 = 14.5 μC.
(e) Yes. Taking R = ∞ corresponds to disconnecting a wire to remove the branch
containing R. In this case |Q|= 0.542 R/R = 0.542 μC.
P28.53
(a)
q = CΔV ( 1 − e− t RC )
−10.0 ⎡( 2.00×106 )( 1.00×10−6 ) ⎤ ⎤
⎣
⎦
q = ( 1.00 × 10−6 F) ( 10.0 V ) ⎡⎢1 − e
= 9.93 μ C
⎣
⎦⎥
(b)
I=
dq ⎛ ΔV
=⎜
dt ⎝ R
⎞ − t RC
⎟e
⎠
⎛ 10.0 V ⎞ −5.00
= 3.37 × 10−8 A = 33.7 nA
I =⎜
⎟e
6
⎝ 2.00 × 10 Ω ⎠
(c)
dU d ⎛ 1 q2 ⎞ ⎛ q ⎞ dq ⎛ q ⎞
= ⎜
= ⎜ ⎟I
⎟=⎜ ⎟
dt
dt ⎝ 2 C ⎠ ⎝ C ⎠ dt ⎝ C ⎠
dU ⎛ 9.93 × 10−6 C ⎞
−8
−7
=⎜
⎟ ( 3.37 × 10 A ) = 3.34 × 10 W = 334 nW
−6
dt ⎝ 1.00 × 10 C V ⎠
(d)
Pbattery = Iε = ( 3.37 × 10−8 A ) ( 10.0 V ) = 3.37 × 10−7 W = 337 nW
The battery power could also be computed as the sum of the instantaneous
powers delivered to the resistor and to the capacitor:
2
−9
2
6
I R + dU/dt = (33.7 × 10 A) 2 × 10 Ω + 334 nW = 337 nW
*P28.54
(a)
We find the resistance intrinsic to the vacuum
cleaner:
( ΔV ) 2
P = I ΔV =
R
2
( ΔV ) ( 120 V )2
R=
=
= 26.9 Ω
P
535 W
with the inexpensive cord, the equivalent
resistance is 0.9 Ω + 26.9 Ω + 0.9 Ω = 28.7 Ω so the
current throughout the circuit is
I=
ε
RTot
=
120 V
= 4.18 A
28.7 Ω
and the cleaner power is
FIG. P28.54
Pcleaner = I ( ΔV )cleaner = I 2 R = ( 4.18 A ) ( 26.9 Ω ) = 470 W
2
In symbols, RTot = R + 2r , I =
ε
R + 2r
and Pcleaner = I 2 R =
ε 2R
( R + 2r )2
12
⎛ ε 2R ⎞
⎛ 26.9 Ω ⎞
R + 2r = ⎜
⎟ = 120 V ⎜
⎟ = 27.2 Ω
⎝ 525 W ⎠
⎝ Pcleaner ⎠
27.2 Ω − 26.9 Ω
ρ l ρl4
r=
= 0.128 Ω =
=
A π d2
2
12
(b)
⎛ 4ρ l ⎞
d=⎜
⎟
⎝ πr ⎠
(c)
12
⎛ 4 ( 1.7 × 10−8 Ω ⋅ m ) ( 15 m ) ⎞
⎟
=⎜
⎜
⎟
π ( 0.128 Ω )
⎝
⎠
R
⎜
2 ⎝ Pcleaner
⎛ 4ρ l ⎞
d=⎜
⎟
⎝ πr ⎠
(a)
= 1.60 mm or more
Unless the extension cord is a superconductor, it is impossible to attain cleaner
power 535 W. To move from 525 W to 532 W will require a lot more copper, as
we show here:
ε⎛
r=
P28.55
12
12
⎞
⎟
⎠
12
−
R 120 V ⎛ 26.9 Ω ⎞
=
⎜
⎟
2
2 ⎝ 532 W ⎠
12
−
⎛ 4 ( 1.7 × 10−8 Ω ⋅ m ) ( 15 m ) ⎞
⎟
=⎜
⎜
⎟
π ( 0.037 9 Ω )
⎝
⎠
26.9 Ω
= 0.037 9 Ω
2
12
= 2.93 mm or more
First determine the resistance of each light bulb: P =
( ΔV )2
( ΔV ) 2
R
( 120 V )2
=
= 240 Ω
P
60.0 W
We obtain the equivalent resistance Req of the network of light
R=
bulbs by identifying series and parallel equivalent resistances:
1
= 240 Ω + 120 Ω = 360 Ω
Req = R1 +
(1 R2 ) + (1 R3 )
FIG. P28.55
The total power dissipated in the 360 Ω is
P =
(b)
( ΔV )2
Req
=
( 120 V )2
360 Ω
= 40.0 W
The current through the network is given by P = I 2 Req :
I=
P
=
Req
40.0 W 1
= A
360 Ω
3
The potential difference across R1 is
⎛1 ⎞
ΔV1 = IR1 = ⎜ A ⎟ ( 240 Ω ) = 80.0 V
⎝3 ⎠
The potential difference ΔV23 across the parallel combination of R2 and R3 is
⎞
1
⎛ 1 ⎞⎛
ΔV23 = IR23 = ⎜ A ⎟ ⎜
⎟ = 40.0 V
⎝ 3 ⎠ ⎜⎝ ( 1 240 Ω ) + ( 1 240 Ω ) ⎟⎠
P28.59
From the hint, the equivalent resistance of.
RT +
That is,
1
= Req
1 RL + 1 Req
RT +
RL Req
RL + Req
= Req
RT RL + RT Req + RL Req = RL Req + Req2
Req2 − RT Req − RT RL = 0
Req =
RT ± RT2 − 4 (1)( − RT RL )
2 (1)
Only the + sign is physical:
Req =
P28.61
1
2
(
4 RT RL + RT2 + RT
For example, if
RT = 1 Ω
And
RL = 20 Ω , Req = 5 Ω
(a)
)
After steady-state conditions have been reached, there is no DC current
through the capacitor.
I R3 = 0 ( steady -state)
Thus, for R3 :
For the other two resistors, the steady-state current is simply determined by
the 9.00-V emf across the 12-k Ω and 15-k Ω resistors in series:
For R1 and R2 :
I ( R1 + R2 ) =
(b)
ε
R1 + R2
=
9.00 V
= 333 μ A ( steady -state)
( 12.0 k Ω + 15.0 k Ω )
After the transient currents have ceased, the potential
difference across C is the same as the potential
difference across R2 ( = IR2 ) because there is no
voltage drop across R3 . Therefore, the charge Q on C is
Q = C ( ΔV ) R = C ( IR2 ) = (10.0 μ F )( 333 μ A )(15.0 kΩ )
2
= 50.0 μ C
FIG. P28.61(b)
(c)
When the switch is opened, the branch containing R1
is no longer part of the circuit. The capacitor discharges
through ( R2 + R3 ) with a time constant
of ( R2 + R3 ) C = ( 15.0 k Ω + 3.00 k Ω ) ( 10.0 μ F) = 0.180 s .
The initial current I i in this discharge circuit is
determined by the initial potential difference across the
capacitor applied to ( R2 + R3 ) in series:
Ii =
( ΔV )C
( R2 + R3 )
=
FIG. P28.61(c)
( 333 μ A ) ( 15.0 k Ω )
IR2
=
= 278 μ A
R
R
+
( 2 3 ) (15.0 k Ω + 3.00 k Ω )
Thus, when the switch is opened, the current through R2 changes
instantaneously from 333 μ A (downward) to 278 μ A (downward) as shown
in the graph. Thereafter, it decays according to
I R2 = I i e− t ( R2 + R3 )C =
(d)
( 278 μ A ) e− t ( 0.180 s) ( for t > 0)
The charge q on the capacitor decays from Qi to
q = Qi e− t ( R2 + R3 )C
Qi
− t 0.180 s)
= Qi e(
5
5 = et 0.180 s
t
180 ms
t = ( 0.180 s) ( ln 5) = 290 ms
ln 5 =
Qi
according to
5