By | December 4, 2020

Direct Current (DC) 

In DC circuits only involve direct current and voltage, that currents and voltages that do not change with time.
The elements in the DC circuit include:

  • Battery
  • Resistance
  • Conductor Wire

battery produces e.m.f to move electrons which ultimately produces an electric current. The term “circuit” is very suitable because in this case a complete electron trajectory must occur – leaving the negative pole and returning to the positive pole. The resistance of the conducting wire is so small that in practice the price is negligible.

The form of resistance (resistor) on the market varies widely, ranging from 0.1 Ω to 10 MΩ or greater. Standard resistors for tolerances of ± 10% are usually rated in multiples of 10 or 0.1 of:

10         12         15         18         22         27         33         39         47        56        68        82

A very simple circuit consisting of a battery with a resistor is shown in figure 1-A. Notice how the two elements are depicted and how to indicate the direction of the current (from the positive pole through the resistor to the negative pole).

fig. 1-A
Direct current circuit:
a) Installation of components and current direction and
b) Addition of switch components and internal resistance.

In Figure 1-B, two other components have been added to the circuit, namely:
i) A switch to break the circuit.
ii) A resistor with the symbol r (lower case) to denote the fact that the battery voltage tends to decrease as the current drawn from the battery increases.

The switch has two conditions:

ON: This condition is commonly referred to as a “short circuit” (shot circuit), which ideally has the following characteristics:values V = 0 for all I (i.e. R = 0) 

OFF: Condition where the current does not flow or so-called as an “open circuit“, ideally it has the following characteristics:values ​​of I = 0 for all V (ie R = ).

In order to analyze it further, it is necessary to understand the basic law of the series which is called Kirchhoff’s law. There are several ways to express Kirchhoff’s law, we try to make it easy to remember:

Figure 2 Simple circuit with three loops 

i) The total current entering a connection point / branch is zero (Law I, called KCL – Kirchhoff curent law ).

∑in = 0                              (1)

The direction of each current is indicated by an arrow, if the current is positive, the current flows in the direction of the arrow, and vice versa. Thus for a series like in Figure 2.2 we can write:

∑in = 0


Negative sign at I1 indicates that the current exits the branch point and if the current enters the branch point is given a positive sign.

ii) In each closed circuit (loop), the amount of voltage drop is zero (Law II, often referred to as KVL – Kirchhoff voltage law)

∑Vn = 0                                  (2)

In Figure 2 using our KVL can write three equations, namely:

For the left loop: – E1 + R3I 3 + R1I1 = 0

For the right loop: – E2 + R2I 2 + R1I1 = 0

For the outer loop : – E1 + R3I 3 – R2I 2 + E2 = 0

Returning to the circuit in Figure 2.1, that all components are passed by current I. According to law II applies:

∑Vn = 0                                                                                       (3)

-E1+I r+ IR  = 0

so the amount of current flowing is

We are interested in


or from equation (3) it is obtained

V = E Ir                                                                   (5)

Equation (5)  shows that the voltage V is the result of the voltage drop due to the load current. The symbol r is called the resistance in the battery. It appears that V is a part (fraction) of E.Such a circuit is commonly referred to as a “voltage divider” (will be discussed further).


Resistors in Series and Parallel Circuits 

This is a basic concept that allows us to quickly simplify relatively complex circuits.

Figure 3.  Resistor in circuit: a) series and b) parallel.

As shown in Figure 3-a, in a series circuit all resistors have the same current. If the current flow of I,we have a

V= I (R1+R2+R3)                    (6)

V / I= R= R1+R2+R3

It appears that for a series circuit, the three resistors can be replaced by a single resistor for R.

In a parallel circuit (figure 3-b), it appears that each resistor gets the same voltage. So





                                                                     G = G1 + G2 + G3                    (8)

where G is commonly referred to as conductance, so G = 1 / R, expressed in units siemen (with the symbol S or mho or Ω-1).


Voltage Divider (Potential Divider

Usually this circuit is used to obtain the desired voltage from a large voltage source. Figure 4 shows a simple form of the voltage divider circuit, that is, it is desirable to get the output voltage V0 which is part of the source voltage V1
by installing two resistors R1 and R2.

Figure 4   Voltage divider circuit


It appears that current i flows through R1 and R2, so that

Vi = Vo+Vs                                  (9)

Vs = i . R1                             (10)

Vo = i . R2                             (11)

Vi = i . R1 + i . R2                  (12)

From equations 10 and 12 it is obtained that

Vo Vs =  R2 / R1                              (.13)

It appears that the input voltage is divided into two parts (Vo, Vs ),

respectively equal to the value of the resistor to which that voltage is applied. 11 and 12 of the equation we get


voltage divider circuit is very important as a basis for understanding the DC circuit or an electronic circuit that involves various components more complicated.


Loaded Voltage Divider 

Figure 5 shows a voltage divider with the load attached to its output terminal, taking a current of  io

 and the voltage drop is Vo.  We will try to find the relationship between  io and Vo . If the current flowing

through R1 is equal to i as shown in the figure, then the current flowing through R2 is i io. We have

ViVo = i x R1                        (15)

Figure 5 Loaded voltage divider circuit.

The voltage at the ends of the load is

Vo = (i io)⋅ R2

Vo = i R2 – ioR2                               (16)

Equations 15 and 16 can be rewritten respectively  as

Vi  x  R2 – VoR2 =  i x  R1 x  R2


VoR1 +ioR1⋅ R2 = i R1⋅ R2

from both of them obtained

Vi  x  R2 – Vx  R2 = Vx  R1 + io x R1 x R2


Vo(R1+ R2) = Vi R2 =  io x R1 x R2 ⋅




VoVo c –  io x⋅ RP                                 (17)

where v0 c is the magnitude of the voltage v0 v without any load, ie when i0 = 0, and

the value  is referred to as the output voltage when the circuit is open(open-circuitoutputvoltage)of






called the “source resistance”, where the value is the same as the resistances R1 and R2 connected in parallel.

Value of v0c or RP depending on the nature of the load, so the effect is V due to the large load can be easily calculated using a simplification of the circuit as shown in Figure 6.

Figure 6 Simplification of the voltage divider circuit

A simple example, suppose that the load installed is a resistance of RL, then the output voltage follows the voltage divider equation, which is

where v0c and RP follows equations 18 and 19, respectively.


Current Divider

Current divider circuit is a voltage divider circuit is not as important, but it should be understood utamannya when we connect the current measuring devices in parallel.

Figure 7 Current divider circuit

In Figure 2.7 it can be seen that v is taken from the resistors R1 and R2, it is clear that

ii = io+ is                      (20)

is = v / R1                    (21)

io = v / R2                   (22)

ii = v / R2 + v / R1     (23)

From equations 21 and 22 we get






where G = 1 / R = conductance.

25 equations indicate that the input current is divided into two parts (0i and Si), each in proportion to the price of the current conductance passed. 22 and 23 of the equation obtained by

io = v / R2





The output current 0i a part (fraction) of the input current.


Thevenin’s Theorem 

The result obtained from the simplification of the circuit is one of the cases of Thevenin’s theorem. In brief, Thevenin’s theorem can be said as follows.

“If a set of voltage source andcircuits are resistorconnected to two output terminals, the 

circuit can be replaced by a series circuit from an open circuit voltage source  Vo /c

and a resistor RP.

Figure 8 shows a circuit network to be connected to a load RL. Combination series Vo /c  and RP in Figure 8-d are

the Thevenin equivalent series.

Figure 8 Schematic formation of the Thevenin equivalent circuit

There are some extreme conditions of the circuit in Figure 8, such as when RL = ∞ and RL = 0 .The value RL = ∞ is in an open circuit, as if it were RL removed from the output terminal, thus obtained an open circuit voltage of V0/C (see figure 8-b). When RL = 0 (figure 8-c) means that the circuit is in a short circuit condition (both ends of the terminal are directly connected )  with a short circuit current I/SC equal to



In some circuits, the calculation of V0/C or  I/SC  likely very difficult to do. The easiest step is to calculate the RP value (resistance value seen from both ends of the output terminal). In this case the RP is calculated looking as if there is no voltage source.

Norton’s Theorem 

This theorem is a series analysis approach which can be briefly said as follows.

“If a set of voltage source and circuits are resistor connected to two output terminals, the

circuit can be replaced by a parallel circuit from a current source short circuit IN and a conductance GN” 

Figure 9 Schematic formation of the Norton equivalent

In Figure 9, the Norton equivalent circuit is depicted with a combination of parallel between an current source  IN and a conductance GN (see Figure 9-d).

If this circuit will be burdened with a load conductance GL,then there are two extreme value is GL = ∞ and  = 0 .
The value of GL = ∞ (or RL = 0 ) is at short circuit condition and short circuit current equal to  IN
While the value GL= 0 (or RL = ∞ ) is in an open circuit condition, which shows that Vo/c is an open circuit voltage. Thus for a Norton equivalent circuit applies :



Problem Exercise

Consider the following circuit:

i) Using Thevenin’s theorem, determine the current flowing in the 3 ohm resistor.

ii) Using the Norton theorem, determine the current flowing in the 3 ohm resistor.


  1. Pingback: BASIC ELECTRONICS |

Comments are closed.