Section 35–9
◆
1057
Series RL and RC Circuits
5. An iron ingot is 1030 °C above room temperature. If it cools exponentially at 3.50% per
minute, find its temperature (above room temperature) after 2.50 h.
6. A certain pulley in a tape drive is rotating at 2550 rev/min. After the power is shut off, its
speed decreases exponentially at a rate of 12.5% per second. Find the pulley’s speed after
5.00 s.
Exponential Growth to an Upper Limit
7. A forging, initially at 0 °C, is placed in a furnace at 865 °C, where its temperature rises
exponentially at the rate of 6.50% per minute. Find its temperature after 25.0 min.
8. If we assume that the compressive strength of concrete increases exponentially with time
to an upper limit of 27.6 MPa, and that the rate of increase is 52.5% per week, find the
strength after 2 weeks.
Motion in a Resisting Medium
9. A 202-N carton is initially at rest. It is then pulled horizontally by a 86.3-N force in the direction of motion, and is resisted by a frictional force equal (in newtons) to four times the carton’s
velocity (in m/s). Show that the differential equation of motion is dv/dt 4.19 0.194v.
10. For the carton in problem 9, find the velocity after 1.25 s.
11. An instrument package is dropped from an airplane. It falls from rest through air whose
resisting force is proportional to the speed of the package. The terminal speed is 47.2 m/s.
Show that the acceleration is given by the differential equation a dv/dt g gv/47.2.
12. Find the speed of the instrument package in problem 11 after 0.50 s.
13. Find the displacement of the instrument package in problem 11 after 1.00 s.
14. A 698-N stone falls from rest from a cliff. If the air resistance is proportional to the square
of the stone’s speed, and the limiting speed of the stone is 38.1 m/s, show that the differential equation of motion is dv/dt g gv2/38.12.
15. Find the time for the velocity of the stone in problem 14 to be 18.3 m/s.
16. A 66.7-N ball is thrown downward from an airplane with a speed of 6.4 m/s. If we assume
the air resistance to be proportional to the ball’s speed, and the limiting speed is 41.1 m/s,
show that the velocity of the ball is given by v 41.1 34.7et/4.19 m/s.
17. Find the time at which the ball in problem 16 is going at a speed of 21.3 m/s.
18. A box falls from rest and encounters air resistance proportional to the cube of the speed.
The limiting speed is 12.5 ft./s. Show that the acceleration is given by the differential equation 60.7 dv/dt 1950 v3.
35–9 Series RL and RC Circuits
Series RL Circuit
Figure 35–11 shows a resistance of R ohms in series with an inductance of L henrys. The switch
can connect these elements either to a battery of voltage E (position 1, charge) or to a short circuit (position 2, discharge). In either case, our objective is to find the current i in the circuit. We
will see that it is composed of two parts: a steady-state current that flows long after the switch
has been thrown, and a transient current that dies down shortly after the switch is thrown.
1
◆◆◆
Example 36: After being in position 2 for a long time, the switch in Fig. 35–11 is thrown into
position 1 at t 0. Write an expression for (a) the current i and (b) the voltage across the inductor.
Solution:
(a) The voltage VL across an inductance L is given by Eq. A87, vL L di/dt. Using
Kirchhoff’s voltage law (Eq. A68) gives
di
Ri L E
dt
R
2
E
i
L
(1) FIGURE 35–11 RL circuit.
1058
Chapter 35
◆
Differential Equations
or
di
dt
R
L
E
L
i (2)
We recognize this as a first-order linear differential equation. Our integrating factor is
e
(RL) dt
3
In electrical problems, we
will use k for the constant
of integration, saving C for
capacitance. Similarly, R here is
resistance, not the integrating
constant.
eRtL
Multiplying Eq. (2) by the integrating factor and integrating, we obtain
E
E
ieRtL eRtL dt eRtL k
L3
R
Dividing by eRtL gives us
E
i keRtL
R
(3)
We now evaluate k by noting that i 0 at t 0, so ke0 k E/R. Substituting into
Eq. (3) yields the following:
Note that the equation for the
current is the same as that for
exponential growth to an upper
limit (Eq. 202), and that the
equation for the voltage across
the inductor is of the same form
as for exponential decay (Eq.
201).
Current in a
Charging Inductor
E
i (1 eRtL)
R
A88
The first term in this expression (ER) is the steady-state current, and the second term (ER)
(eRtL) is the transient current.
(b) From (1), the voltage across the inductor is
di
vL L E Ri
dt
Using Eq. A88, we see that Ri E EeRtL. Then
vL E E EeRtL
Thus,
Voltage across a
Charging Inductor
vL EeRtL
A90
Series RC Circuit
We now analyze the RC circuit as we did the RL circuit.
◆◆◆
1
Example 37: A fully charged capacitor (Fig. 35–12) is discharged by throwing the switch
from 1 to 2 at t 0. Write an expression for (a) the voltage across the capacitor and (b) the
current i.
R
2
E
Solution:
i
C
FIGURE 35–12 RC circuit.
(a) By Kirchhoff’s law, the voltage vR across the resistor at any instant must be equal to the
voltage v across the capacitor, but of opposite sign. Further, the current through the resistor, vR R, or vR, must be equal to the current C dvdt in the capacitor.
v
dv
C
R
dt
Section 35–9
◆
1059
Series RL and RC Circuits
Separating variables and integrating gives
dt
RC
t
ln v k
RC
But at t 0 the voltage across the capacitor is the battery voltage E, so k ln E. Substituting,
we obtain
v
t
ln v ln E ln E
RC
Or, in exponential form, v/E etRC, which is also expressed as follows:
dv
v
Voltage across a
Discharging Capacitor
v EetRC
A85
(b) We get the current through the resistor (and the capacitor) by dividing the voltage v by R.
Current in a
Discharging Capacitor
E
i etRC
R
Equations A83 and A85 are both
for exponential decay.
A83
◆◆◆
◆◆◆ Example 38: For the circuit of Fig. 35–12, R 1540 , C 125 F, and E 115 V. If
the switch is thrown from position 1 to position 2 at t 0, find the current and the voltage across
the capacitor at t 60 ms.
Solution: We first compute 1RC.
1
1
5.19
RC 1540 125 106
Then, from Eqs. A83 and A85,
115
E
i etRC e5.19t
R
1540
and
v EetRC 115e5.19t
At t 0.060 s, e5.19t 0.732, so
115
i (0.732) 0.0547 A 54.7 mA
1540
and
v 115(0.732) 84.2 V
◆◆◆
Alternating Source
We now consider the case where the RL circuit or RC circuit is connected to an alternating
rather than a direct source of voltage.
E sin ωt
R
L
Example 39: A switch (Fig. 35–13) is closed at t 0, thus applying an alternating voltage FIGURE 35–13 RL circuit
with ac source.
of amplitude E to a resistor and an inductor in series. Write an expression for the current.
◆◆◆
1060
Chapter 35
◆
Differential Equations
Solution: By Kirchhoff’s voltage law, Ri L didt E sin t, or
di
dt
R
L
E
L
i sin t
This is a first-order linear differential equation. Our integrating factor is e
Thus,
E
ieRtL eRtL sin t dt
L3
RL dt
3
eRtL.
eRtL
E
R
p sin t cos t q k
L (R2/L2 2) L
by Rule 41. We now divide through by eRtL and after some manipulation get
Z=
R2
+
R sin t L cos t
i E · keRtL
R2 2L2
ω 2L2
XL = ωL
R
From the impedance triangle (Fig. 35–14), we see that R2 2L2 Z2, the square of the
impedance. Further, by Ohm’s law for ac, E/Z I, the amplitude of the current wave.
Thus,
L
E R
i p sin t cos t q keRtL
Z Z
Z
FIGURE 35–14 Impedance
triangle.
L
R
I p sin t cos t q keRtL
Z
Z
Again from the impedance triangle, R/Z cos and L/Z sin . Substituting yields
i I(sin t cos cos t sin ) keRtL
I sin (t ) keRtL
which we get by means of the trigonometric identity for the sine of the difference of two angles
(Eq. 167). Evaluating k, we note that i 0 when t 0, so
IXL
k I sin () I sin Z
where, from the impedance triangle, sin XL Z. Substituting, we obtain
IXL
i I sin(t ) eRtL
Z
steady-state transient
current
current
Our current thus has two parts: (1) a steady-state alternating current of magnitude I, out of phase
with the applied voltage by an angle ; and (2) a transient current with an initial value of IXL Z,
◆◆◆
which decays exponentially.
Exercise 9
◆
Series RL and RC Circuits
Series RL Circuit
1. If the inductor in Fig. 35–11 is discharged by throwing the switch from position 1 to 2,
show that the current decays exponentially according to the function i (E/R)eRtL.
2. The voltage across an inductor is equal to L di/dt. Show that the magnitude of the voltage across the inductance in problem 1 decays exponentially according to the function
v EeRtL.
Section 35–9
◆
Series RL and RC Circuits
3. We showed that when the switch in Fig. 35–11 is thrown from 2 to 1 (charging), the current
grows exponentially to an upper limit and is given by i (E/R)(1 eRt/L). Show that the
voltage across the inductance (L di/dt) decays exponentially and is given by v EeRt/L.
4. For the circuit of Fig. 35–11, R 382 , L 4.75 H, and E 125 V. If the switch is
thrown from 2 to 1 (charging), find the current and the voltage across the inductance at t
2.00 ms.
Series RC Circuit
5. The voltage v across the capacitor in Fig. 35–12 during charging is described by the differential equation (E v)/R C dv/dt. Solve this differential equation to show that the
voltage is given by v E(1 et/RC). (Hint: Write the given equation in the form of a firstorder linear differential equation, and solve using Eq. 427.)
6. Show that in problem 5 the current through the resistor (and hence through the capacitor)
is given by i (E/R)et/RC.
7. For the circuit of Fig. 35–12, R 538 , C 525 F, and E 125 V. If the switch is
thrown from 2 to 1 (charging), find the current and the voltage across the capacitor at t 2.00 ms.
Circuits in Which R, L, and C Are Not Constant
8. For the circuit of Fig. 35–11, L 2.00 H, E 60.0 V, and the resistance decreases with
time according to the expression
R 4.00/(t 1)
Show that the current i is given by
i 10(t 1) 10(t 1)2
9. For the circuit in problem 8, find the current at t 1.55 ms.
10. For the circuit of Fig. 35–11, R 10.0 , E 100 V, and the inductance varies with
time according to the expression L 5.00t 2.00. Show that the current i is given by the
expression i 10.0 40.0/(5.00t 2.00)2.
11. For the circuit in problem 10, find the current at t 4.82 ms.
12. For the circuit of Fig. 35–11, E 300 V, the resistance varies with time according to the expression R 4.00t, and the inductance varies with time according to the expression L t2 4.00.
Show that the current i as a function of time is given by i 100t (t2 12.0)/(t2 4.00)2.
13. For the circuit in problem 12, find the current at t 1.85 ms.
14. For the circuit of Fig. 35–12, C 2.55 F, E 625 V, and the resistance varies with current according to the expression R 493 372i. Show that the differential equation for
current is di/dt 1.51i di/dt 795i 0.
Series RL or RC Circuit with Alternating Current
15. For the circuit of Fig. 35–13, R 233 , L 5.82 H, and e 58.0 sin 377t V. If the switch
is closed when e is zero and increasing, show that the current is given by i 26.3 sin(377t 83.9°) 26.1e40t mA.
16. For the circuit in problem 15, find the current at t 2.00 ms.
17. For the circuit in Fig. 35–12, the applied voltage is alternating and is given by e E sin t.
If the switch is thrown from 2 to 1 (charging) when e is zero and increasing, show that the
current is given by
i (E/Z)[sin(t ) et/RC sin ]
In your derivation, follow the steps used for the RL circuit with an ac source.
18. For the circuit in problem 17, R 837 , C 2.96 F, and e 58.0 sin 377t. Find the
current at t 1.00 ms.
1061
1062
Chapter 35
35–10
◆
Differential Equations
Second-Order Differential Equations
The General Second-Order Linear Differential Equation
A linear differential equation of second order can be written in the form
Py Qy Ry S
where P, Q, R, and S are constants or functions of x.
A second-order linear differential equation with constant coefficients is one where P, Q, and
R are constants, although S can be a function of x, such as in the following equation:
We will show a numerical method
for approximately solving secondorder differential equations in the
following chapter.
Form of Second-Order
Linear DE, Right Side
Not Zero
ay by cy f (x)
433
where a, b, and c are constants. This is the type of equation we will solve in this chapter.
Equation 433 is said to be homogeneous if f(x) 0, and it is called nonhomogeneous if
f(x) is not zero. We will, instead, usually refer to these equations as “right-hand side zero” and
“right-hand side not zero.”
We’ll usually use the more
familiar y notation rather than
the D operator.
Operator Notation
Differential equations are often written using the D operator that we first introduced in Chapter
27, where
Dy y
D2y y
D3y y
etc.
Thus, Eq. 433 can be written
aD2y bDy cy f(x)
Second-Order Differential Equations with Variables Separable
We develop methods for solving the general second-order equation in Sec. 35–11. However,
simple differential equations of second order that are lacking a first derivative term can be
solved by separation of variables, as in the following example.
◆◆◆ Example 40: Solve the equation y 3 cos x (where x is in radians) if y 1 at the point
(2, 1).
Solution: Replacing y by d(y)/dx and multiplying both sides by dx to separate variables, we
have
d(y) 3 cos x dx
Notice that we have to integrate
twice to solve a second-order
DE and that two constants of
integration have to be evaluated.
Integrating gives us
y 3 sin x C1
Since y 1 when x 2 rad, C1 1 3 sin 2 1.73, so
y 3 sin x 1.73
or dy 3 sin x dx 1.73 dx. Integrating again, we have
y 3 cos x 1.73x C2
At the point (2, 1), C2 1 3 cos 2 1.73(2) 3.21. Our solution is then
y 3 cos x 1.73x 3.21
◆◆◆