Supplementary Application 3:
Electric Networks
We can analyze an electric network by using a combination of differential and algebraic
equations. To understand the analysis it will be enough for our purposes to become
familiar with the following ideas. An LRC network consists of elements called inductors
(with inductance Lj measured in henrys), resistors (with resistance Rj measured in ohms),
and capacitors (with capacitance Cj measured in farads.) These elements are joined by
conductors in such a way that any two points of the network can be included in a closed
loop, or circuit, contained in the network. (Such circuits are considered in isolation
from one another in Chapter 4, Section 2.) The numbers Lj , Rj , and Cj describe the
essential character of a network. In addition a network typically contains voltage sources
characterized by voltages Ej together with the direction in which each source Ej would
by itself cause current to flow. A point of the network at which entering current can flow
out on more than one conductor is called a junction. Two junctions are shown in Figures
1 (a) and (b), and four in Figure 1 (c). The segment of a network joining two successive
junctions is called a branch. Six branches are shown in Figure 1 (c), while there are
only three each in (a) and (b). The branches are chosen so each will contain at most one
voltage source and at most one of each type of network element. In principle there is some
resistance, however small, present in every branch, but it may sometimes be negligibly
small.
Figure 1
The problem usually posed is to find the current flowing in the jth branch at time
t; we’ll denote this current by Ij (t), measured in amperes. We follow the convention that
if current is flowing in an arbitrary fixed direction in the jth branch then Ij > 0, while
flow in the opposite direction corresponds to Ij < 0. In analyzing a network we assign
an arbitrary direction, designated “positive,” to each branch and indicate our choices by
arrows in the network diagram.
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Supplementary Application 3
Relations between the currents Ij (t) sufficient to determine their values can be derived
from initial conditions together with the two Kirchhoff laws:
Junction law. The sum of currents directed toward a junction equals the sum of the
currents directed away from it.
Loop law. The sum of voltage differences across the elements in a closed loop equals
the sum of the source voltages in the loop.
The voltage differences in a loop, referred to in the loop law, are caused by the presence
of one or more of inductance, resistance or capacitance in the loop, and are computed for
the current Ij in the branch containing such a network element as follows.
Voltage difference Lj dIj /dt is due to inductance Lj .
Voltage difference Rj Ij is due to resistance Rj . (Ohm’s law)
Voltage difference Qj /Cj is due to capacitance Cj , where Qj is the charge on the
capacitor, and charge is related to current byIj = dQj /dt.
In computing the sum of voltage differences in a loop, we assign arbitrarily a fixed direction
of traversal to the loop, either clockwise or counterclockwise, and attach a minus sign to
a difference for which this direction of traversal is counter to our previously assigned
direction of the branch that contains the element causing the difference. A voltage source
is designated positive if by itself it would cause current to flow in the direction of traversal
of a loop and negative otherwise.
Let us consider first what happens in a network that doesn’t contain any capacitors;
this gives rise immediately to a system of equations to be satisfied by the currents Ij (t). If
the network doesn’t contain any inductors either, the system will consist of purely algebraic
relations between the currents Ij (t) and whatever voltage sources the network contains. If
one or more inductors are also present the system will consist of a combination of algebraic
equations and of differential equations of order one. The following example is of this type.
Example 1 Figure 1(a) shows a network that contains just resistors and inductors in
addition to a single voltage source. Applying the junction law gives the same equation
regardless of which of the two junctions it is applied to. Either way we get
I1 + I2 = I3
Suppose that the voltage source causes current to flow in the direction shown for I3 when
E > 0. The loop-law applied to the left-hand loop then gives
L1 I˙1 + R1 I1 + R3 I3 = E.
Electric Networks
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From the right-hand loop we get an equation of the same form relating I2 and I3 , with L2
replacing L1 and R2 replacing R1 :
L2 I˙2 + R2 I2 + R3 I3 = E.
Replacing I3 by I1 + I2 gives two differential equations for I1 and I2 :
L1 I˙1 + (R1 + R3 )I1 + R3 I2 =E
L2 I˙2 + (R2 + R3 )I2 + R3 I1 =E.
Suppose that L1 = L2 = 0.1 and that R1 = 10, R2 = 20 and R3 = 30. Suppose also that at
the time we first observe the network the voltage source has been switched off and replaced
by a conductor. Then E = 0, and our differential equations can be written
I˙1 + 400I1 + 300I2 =0
300I1 + I˙2 + 500I2 =0.
In operator form these equations are
(D + 400)I1 + 300I2 =0
300I1 + (D + 500)I2 =0.
The elimination method leads to the characteristic equation
r2 + 900r + 110000 = 0,
with roots (to the nearest integer) r1 = −754, r2 = −146. An approximation to the general
solution then has the form
I1 (t) ≈150c1 e−754t + 150c2 e−146t
I2 (t) ≈177c1 e−754t − 127c2 e−146t .
Initial values I1 (0) and I2 (0) can be used to determine acceptable values for c1 and c2 .
We’ve already seen the relation dQ/dt = I that relates the current in a branch to
the charge on a capacitor in the branch. Equations containing a charge Qj are usually
differentiated once with respect to t in order to eliminate Qj and get equations entirely in
terms of currents Ij . If this is done, the equation for a loop containing both an inductor
and a capacitor will become a second-order differential equation for Ij because it already
contains a term of the form Lj dIj /dt, whose derivative is Lj d2 Ij /dt2 .
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Supplementary Application 3
Example 2 The network shown in Figure 1(b) contains circuit elements of all three types:
R, L and C. Applying the junction law to either junction gives the same equation:
I1 + I2 = I3 .
Suppose that the voltage source causes current to flow in the direction shown for I3 when
E > 0. Applying the loop law to the left-hand loop then gives
L1 I˙1 + R1 I1 + R3 I3 = E.
If Q2 (t) represents the charge on the capacitor, so that Q̇2 = I2 , the right hand loop yields
L2 I˙2 + Q2 /C2 + R3 I3 = E.
If we differentiate this last equation with respect to t and replace Q̇2 by I2 , we get
L2 I¨2 + (1/C2 )I2 + R3 I˙3 = Ė.
Finally we can replace I3 by I1 + I2 to get a pair of differential equations for I1 and I2 :
L1 I˙1 + (R1 + R3 )I1 + R3 I2 = E
L2 I¨2 + R3 I˙2 + (1/C2 )I2 + R3 I˙1 = Ė.
The form of these two differential equations suggests that specifying initial values for I1 , I2
and I˙2 will be enough to determine I1 (t) and I2 (t), and hence also their sum I3 (t). (Note
that I˙2 (0) is determined by the equation for the right-hand loop if we know Q2 (0), the
initial charge on the capacitor, as well as E(0) and I3 (0).) If E represented a constant
voltage source of size V0 that caused current to flow up in the central branch, then E
should be replaced by −V0 , because the upward direction is counter to the way the two
loops were traversed. For a downward directed source, we would set E = V0 . In either
case Ė = 0, because E is constant. For a variable voltage source, similar remarks apply.
For example, if E(t) = sin t, then Ė(t) = cos t, and in the time interval between 0 and π
the source would be causing current to flow down in the central branch, followed by an
upward flow in the next time interval of length π, etc.
EXERCISES
The three networks shown below differ only in the elements they contain; each one results
from the one to its left by including a new element. In these diagrams let the symbol
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E stand for a constant voltage source of E volts that would cause current to flow down
relative to the diagram. (Such a source could be provided by a battery with its (+) terminal
attached to the junction below it and its (-) terminal attached to the resistor above it.)
1. In the network shown in Figure 2 (a), let R1 = 5, R2 = 10, R3 = 15 and E = 110.
(a) Find a system of equations that determines the currents in the three branches and
solve the system.
(b) What happens to the solution in part (a) if the voltage source is applied upward instead
of downward?
(c) What happens to the solution if R3 = 0 instead of 15?
(d) What happens to the solution if R1 = R3 = 0?
Figure 2
2. In the network shown in Figure 2 (b) let R1 = 5, R2 = 10, R3 = 15, L1 = 0.1, and
E = 110.
(a) Find a system of equations that determines the currents in the three branches.
(b) Find the general solution of the system of equations found in part (a).
(c) Determine the constant in the general solution by imposing the condition that at time
t = 0 a current of 3 amperes is flowing upward in the left branch.
(d) Find the limit as t tends to infinity of the solutions found in part (c).
3. In the network shown in Figure 2(c) let R1 = 5, R2 = 10, R3 = 15, L1 = 0.1, C2 =
1/250 = 0.004, E = const.
(a) Find a system of equations that determines the currents in the three branches.
(b) Find the general solution of the system of equations found in part (a).
(c) Determine the constants in the general solution by imposing the condition that at time
t = 0 currents of 2 and 3 amperes respectively are flowing upward in the right and left
branches and that the initial charge on the capacitor is Q2 (0) = 5.
4. In the network in Figure 2(c) let R1 = 5, R2 = 10, R3 = 15, L1 = 0.1, C2 = 1/1250 =
8 × 10−4 , E = const.
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Supplementary Application 3
(a) Find a system of equations that determines the currents in the three branches.
(b) Find the general solution of the system of equations found in part (a), and show that
they exhibit damped oscillation.
(c) Determine the constants in the general solution by imposing the condition that at time
t = 0 currents of 2 and 3 amperes respectively are flowing upward in the right and left
branches and that the initial charge on the capacitor is Q2 (0) = 5.
5. Find a system of equations for the currents in the six branches of the network shown
in Figure 1(c), assuming general constant values for the voltages and also for the characteristics of the circuit elements. Note that positive current directions have not been
assigned.
6. Suppose two completely independent LRC circuits get joined by a single conductor
connecting a point in each circuit. Explain why the resulting configuration is not a network
in the sense of our definition.