May 1973 Popular Electronics
Table of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from
Popular Electronics,
published October 1954  April 1985. All copyrights are hereby acknowledged.

A series of three articles
appeared in 1973 issues of Popular Electronics magazine that conducted
a highlevel review  or introduction if you've never seen it before  of
DC circuit analysis. In this first installment from the May issue, Professor Arthur Seidman,
of the Pratt Institute,
covers a variety of subjects starting with direct current (DC) circuit theory.
Ideal current and voltage sources, units and notations, Ohm's law, Kirchhoff's
law, resistors, capacitor and inductor charge and discharge curves, series
and parallel circuits, power calculations, conductance, and other good stuff
is covered. There is even (gasp) a bit of calculus presented.
Do You Know Your DC Circuits?
Part 1 of 1 3Part Series Covering
DC Circuit Analysis
By Arthur H. Seidman, Prof. of Elect Eng., Pratt Institute
Editor's Note: Whether you are an experienced designer and finished your
formal education years ago or are just starting your first courses in electronics,
here is an excellent opportunity to learn the fundamentals of DC circuit theory.
Start now, and don't miss the second and third parts in succeeding issues.
Other subjects to be covered in these series include transistors and diodes.
1. Passive Elements
A. Definition
An element is passive if it is only capable of accepting electrical energy.
(A battery, for example, provides electrical energy and is therefore called
an active element.) Examples of passive elements are resistors (R), capacitors
(C), and inductors (L).
B. Ohm's Law
The current i (amps) in a resistor R (ohms) is equal to the voltage v (volts)
across the resistor divided by the value of the resistance: i = v/R. This
is the basic statement of Ohm's law. By algebraic manipulation, one can also
write v = iR and R = v/i.
C. Voltage and Current Relations for L and C
(1) The voltage across an inductor (L) is equal to the value of inductance
(henrys) multiplied by the rate of change of current with respect to time
(amperes/second): v = L(di/dt). If i is a direct current, it does not change
with time; therefore di/dt = 0 and v = 0. (2) The current in a capacitor (C)
is equal to the value of the capacitance (farads) multiplied by the rate of
change of voltage across the capacitor with respect to time (volts/second):
i = C(dv/dt). If v is a direct voltage, it does not change with time; therefore,
dv/dt = 0 and i = 0. This demonstrates that a capacitor blocks the flow of
direct current. (3) The charge q (coulombs) stored in a capacitor is q = vC.
2. Linear Elements
A. Definition
A linear element is one that, if its input is increased by a given amount,
responds with a proportional increase. For example, if the voltage across
a linear resistor is doubled, the current flowing in the resistor is also
doubled. A nonlinear resistor is shown in Fig. 1. Note that the value
of the resistor, R (v), depends on the voltage across it. (In this article
we are concerned only with linear elements.)
B. Linear Circuits
A linear circuit contains only linear elements.
3. Notation
For DC quantities, use uppercase letters for voltage (V), current (I),
energy (W), and power (P). For timevarying quantities, use lowercase letters.
4. Ideal Sources
A. Definition
(1) An ideal voltage source is one whose voltage, V" is constant
regardless of the current it supplies. (2) An ideal current source is one
whose current, I, is constant regardless of the voltage across the element
to which it supplies current.
B. Symbols
Symbols for ideal voltage and current sources are shown in Fig. 2.
5. Unit Step Function
A. Definition
A unit step function, u_{1} (t), is equal to 0 for t less than
0 and 1 for t greater than 0 (Fig. 3). This concept is useful in describing,
for example, the application of a direct voltage (or direct current) to a
circuit. If we let v = Vu_{1}(t), at t less than 0, v = 0 and at
t greater than 0, v = V. If i = Iu_{1},(t ), i = 0 for t less than
0 and i = 1 for t greater than 0.
6. Circuits Containing R & C and R & L Elements
A. RC Circuits
To a direct current, a capacitor acts initially (time t = 0) like a short
circuit. (1) The product of R and C is called the time constant, T, of the
circuit: T = RC. (2) After a time of approximately 5T, the circuit is considered
to be in steady state, and the capacitor acts like an open circuit. (3) Steady
state is symbolized by t = ∞.
Ex. 1. For the circuit in Fig. 4, determine 1 at (a) t = 0, I(0) and
(b) in steady state, 1(∞). (c) What is the time constant of the circuit?
Sol. (a) I(0) = 10/100 = 0.1 A. (b) I(∞) = 0. (c) T = RC = 100 X 10
X 10^{6} = 10^{3} s = 1 ms.
Ex. 2. For the circuit in Fig. 4, find the voltage across the capacitor,
V_{C}, at (a) t = 0 and (b) t = ∞. Sol. (a) Because the capacitor
acts like a short circuit at t = 0, V_{C}(0) = 0. (b) Because the
capacitor acts like an open circuit in steady state, V_{C}(∞)
= 10 V.
B. Functions of Time
For times between t = 0 and t = ∞, the current and voltage as functions
of time for the circuit in Fig. 4 may be expressed by:
i(t) = (V/R) ε^{t/RC} and vc(t) = V (1  ε^{t/RC}).
C. RL Circuits
To a direct current, an inductor acts like an open circuit at t = 0 and
as a short circuit at t = ∞. The time constant is T = L/R.
Ex. 3. Referring to Fig. 5, determine the current I at (a) t = 0 and
(b) t = ∞. (c) What is the time constant for the circuit? Sol. (a) I(0)
= 0. (b) I(∞) = V/R = 10/10 = 1 A. (c) T = L/R = 0.1/10 = 0.01 s = 10
ms.
Ex. 4. For the circuit in Fig. 5, find the voltage across the inductor,
at (a) t = 0 and (b) t = ∞. Sol. (a) Because the inductor acts like
an open circuit at t = 0, V_{L}(0) = 10 V. (b) Because the inductor
acts like a short circuit in steady state, V_{L}(∞) = 0.
D. Functions of Time
For times between t = 0 and t = ∞, the current and voltage as functions
of time for the circuit in Fig. 5 may be expressed by: i(t) = (V/R) (i
 ε^{tR/L}) and v_{L}(t) = V ε^{tR/L}.
7. Energy and Power
A. Definitions
(1) Energy may be defined as the ability to do work. In the mks (meter,
kilogram, second) system of units, the unit of energy is the newtonmeter
or joule. One joule is equal to one wattsecond. (2) Power is the rate of
doing work or the rateofchange. of energy with respect to time. Its unit
is joules/second or watts. (3) Energy, w, is equal to the product of power,.
p, and time, t: w = pt. (4) For electrical circuits, p = vi. Since i = v/R
and v = iR, we can write p = v^{2}/R = i^{2}R. (5) Only resistors
are capable of dissipating energy; capacitors and inductors store energy.
B. Energy Stored in Capacitor
The energy stored in a capacitor is w_{C} = 1/2Cv^{2},
where v is the voltage across the capacitor.
C. Energy Stored in Inductor
The energy stored in an inductor is w_{L} =:= 1/2Li^{2},
where i is the current flowing in the inductor.
Ex. 5. A 200watt light bulb is powered by a 120volt DC source. Find (a)
the current required by the bulb and (b) the energy consumed in lighting the
bulb for 2 hours. Sol. (a) I = P/V = 200/120 = 1.67 A. (b) W = Pt = 200 X
2 X 3600 = 1.44 X 10^{6} joules.
Ex. 6. Determine the energy stored in (a) a 1microfarad capacitor across
which is 10 volts and (b) a 10millihenry inductor whose current is 4 A. Sol.
(a) W_{C} = 1/2 X 10^{6} (10)^{2} = 0.5 X 10^{4}
joules. (b) w_{L} = 1/2 X 10 X 10^{3} (4)^{2} = 0.08
joules.
Note: The circuit laws, theorems, and techniques reviewed here also apply,
in general, to AC circuits.
8. Kirchhoff's Laws
A. Kirchhoff's laws provide a formal method of finding
currents and voltages in a circuit, regardless of its complexity. The laws
apply to circuits energized by de, ac, or timevarying voltage and currents.
B. Voltage Law
The voltage law states that the algebraic sum of voltages
around a closed path is equal to zero. Mathematically, ∑v(t) = 0, where ∑
is the Greek symbol for sum.
C. Series Circuit
A series
circuit is a network where the same current flows in each element. An example
of a series circuit is shown in Fig. 6.
Ex. 7. Using Kirchhoff's voltage law, find (a) the current in and (b) the
voltage across the 10ohm resistor in the series circuit in Fig. 6. Sol.
It is necessary to distinguish the direction of current flow and the polarity
of voltage sources in the application of Kirchhoff's voltage law. The convention
used here is: (1) A minus sign precedes a voltage quantity in going around
a closed path from a low to a high potential or against the direction of indicated
current flow. (2) A plus sign precedes a voltage quantity in going around
a closed path from a high to a low potential or with the direction of indicated
current flow. (3) The assumed direction of current flow is arbitrary. (a)
Referring to Fig. 6, assume that we start at point A and go around the
closed path clockwise (with the direction of indicated current flow). Then,
according to the voltage law and the conventions we have adopted,  100 +
(25 + 10 +15)I = 0 or I = 100/50 = 2 A. (b) From Ohm's law, the voltage across
the resistor is V = IR = 2(10) = 20 V.
Ex. 8. Figure 7 shows a twomesh or twoloop
network with the assumed clockwise direction of current flow in each mesh
indicated. Write the necessary equations to solve for the two currents. Sol.
For the first mesh, beginning at point A, the voltages are V_{1}
+ (R_{1} + R_{2})I_{1}  R_{2}I_{2}
= 0 (Eq 1). For the second mesh, starting at point B, R_{2}I_{1}
+ (R_{2} + R_{3})I_{2} + V_{2} = 0 or V_{2}
= R_{2}I_{1} + (R_{2} + R_{3})I_{2}
(Eq 2). Equations (1) and (2) constitute a pair of simultaneous equations.
Terms R_{1} + R_{2} and R_{2} + R_{3} are
the self resistance of meshes 1 and 2, respectively; term R_{2} is
the mutual resistance linking the two meshes.
Ex. 9. If, in Ex. 8, R_{1} = 1 ohm, R_{2} = 2 ohms, R_{3}
= 4 ohms, and V_{1} = V_{2} = 6 V, solve for I_{1},
I_{2} and the voltage across each resistor. Sol. Substituting the
given values in Eqs (1) and (2), we get 6 = 3I_{1}  2I_{2}
(Eq 3) and 6 = 2I_{1} + 6I_{2} (Eq 4). Equations (3) and
(4) may be solved for I_{1} and I_{2} in a number of ways.
If, for example, we multiply Eq (3) by 3 and add the result to Eq (4), we
have 12 = 7I_{1} or I_{1} = 12/7 A. Substituting this value
in Eq (4) and solving, I_{2} = 3/7 A. The negative sign denotes that
I_{2} actually flows in a direction opposite to that assumed in Fig. 7.
The voltage across R_{1} is V = (12/7) 1 = 12/7 V; across R_{2},
V = (12/7 + 3/7)2 = 30/7 V; and across R_{3}, V = (3/7)4 = 12/7
V.
D) Current Law
Kirchhoff's current law states that the algebraic sum of currents leaving
a node is equal to zero, or ∑i(t) = 0. A node is a junction of two or
more elements. If currents flowing toward a node are taken as negative and
currents flowing away as positive, then, for example, i_{1} + i_{2}
+i_{3} = 0 in Fig. 8.
E) Parallel Circuits
A parallel circuit is a network where the voltage across each element is
the same (Fig. 9).
Ex. 10. For the parallel circuit in Fig. 9, determine (a) the voltage
across each resistor and (b) the current in each. Sol. (a) Applying Kirchhoff's
current law at node A, we have: 6 + V/2 + V/4 = 0 or 6 = V(2 + 1)/4 = 3V/4;
and V = 8V. (b) I_{2} = 8/2 = 4 A; I_{4} = 8/4 = 2 A.
F) Conductance
Conductance, G, is the reciprocal of resistance; thus G = 1/R. The unit
for conductance is the mho, which is ohm spelled backwards.
Ex. 11. Referring to Fig. 10, write the necessary equations to solve
for V_{1} and V_{2}. Sol. From Fig. 10, V_{1}
is taken between nodes 1 and N; with V_{2} between nodes 2 and N.
Node N is a common node and nodes 1 and 2 are independent nodes. Because we
have two independent nodes, two nodal equations are necessary. With the assumed
currents as shown, Kirchhoff's current law for node 1 gives:  I_{1}
+ I_{2} + I_{3} = 0 (Eq. 5). At node 2: I_{3} + I_{4}
= 0 (Eq 6). Expressing Eqs (5) and (6) in terms of voltages and resistance,
we have (V  V_{1})/R_{1} + V_{1}/R_{2} +
(V_{1}  V_{2})R_{3} = 0, or V/R_{1} = V_{1}(1/R_{1}
+1/R_{2} + 1/R_{1})  V_{2}/R_{3} (Eq 7);
and  (V_{1}  V_{2})/R_{3} + V_{2}/R_{4}=
0 or V_{1}/R_{3} + V_{2}(1/R_{3} + 1/R_{4})
= 0 (Eq 8). Equations (7) and (8) can be expressed in terms of conductances:
thus, G_{1}V = (G_{1} + G_{2} + G_{3})V_{1}
 G_{3V2} (Eq 9) and G_{3}V_{1} + (G_{3}
+ G_{4})V_{2} = 0 (Eq 10). Terms G_{1} + G_{2}
+ G_{3} and G_{3} + G_{4} are the self conductance
of nodes 1 and 2, respectively; and G_{3} is the mutual conductance
between nodes 1 and 2.
Ex. 12. For the network in Fig. 11, determine V_{1} and V_{2}.
Sol. At node 1, 10 + 0.5V_{1} + 0.5 (V_{1}  V_{2})
= 0, or 10 = V_{1}  0.5V_{2} (Eq 11). At node 2,  0.5(V_{1}
 V_{2}) + 0.5V_{2} = 0, or 0.5V_{1} +V_{2}
= 0 (Eq 12). From Eq (12), V_{1} = 2V_{2}. Substituting this
value in Eq 11, we have 10 = 2V_{2}  0.5V_{2} = 1.5V_{2}.
Solving, V_{2} = 10/1.5 = 6.7 V and V_{1} = 2V_{2}
= 2(6.7) = 13.4 V.
Fig. 11
(To be continued)
Posted August 21, 2024 (updated from original
post on 2/16/2018)
