Introduction
Circuit elements are produced with standard size terminals to facilitate their assembly. One way to mount circuits, without having to weld (left figure), is with a test board ( breadboard ). To construct longer lasting circuits, you can use a circuit board ( stripboard ), which is a plate of insulating material with holes and parallel copper tracks on one side (right figure); contact between the various components is achieved by inserting the terminals into holes that are on the same track as the test plate, but soldering the terminals on copper. They also can be constructed more compact circuits using plates circuit printed (PCB). A PCB is similar to a circuit board, but the copper tracks and the holes are designed to measure for each particular circuit.
5.1. Circuit diagrams
A DC circuit (DC stands for Direct Current), is a circuit in which all voltage sources have constant electromotive force and the resistances are all constant. If the circuit includes capacitors, the current may vary over time (transient response of the circuit), but after some time the charge and voltage on the capacitors reach constant values.
This chapter explains how to calculate the initial and final values of currents and charges and in the chapter on signal processing studies the analysis of the transient response of the DC circuits. To analyze circuits is convenient to use the simplified diagrams. A circuit diagram of the assembly example is used to charge a capacitor and then to observe how the potential difference decreases when the capacitor is discharged through a voltmeter. The circuit diagram is shown in Figure 5.1 . The cell binds to the capacitor for a time until it is charged and is therefore off. The action on and off the stack is represented in the circuit diagram for the switch, which will be closed while the capacitor is charged.
The voltmeter is represented in the diagram by its internal resistance . It is generally assumed that the voltmeter does not interfere with the circuit and it is represented only by arrows with positive and negative signs, which indicate the points where the positive and negative terminals of the voltmeter are connected. In this case the voltmeter resistance is important and that is why it has been included in the diagram. An ideal voltmeter would have infinite resistance, which would not allow the condenser unloaded, remaining its constant potential difference. In a real voltmeter, the charge on capacitor produces a current through the voltmeter, which decreases the load and hence the potential difference.
5.2. Circuit laws
The analysis of a circuit consists in calculating the current or potential difference in each resistance and the charge or potential difference in each capacitor. With these quantities one can also determine the power being dissipated in the resistances and the energy stored in the capacitors. To analyze the circuit it is convenient to use two general rules called Kirchhoff laws .
The first law, called the nodes law or currents law, states that at any point in a circuit where there is separation of the current (a node) the sum of the currents entering the node must be equal to the sum of currents leaving it. For example, in the node depicted in Figure 5.2, there is one current entering the node and two currents and leaving the node. The current law implies:
This law is valid when the currents are stationary; in particular, when the density of the conducting charges cloud remains constant within the conductor, without accumulation of charges on any point. In that case, the entire charge entering one end of the conductor, per unit of time, will leave the through the other end.
The second law called the loops law , or voltages law, states that the sum of the potential differences in any closed path (a loop) in a circuit is always zero.
For example, in the circuit of Figure 5.3, one can identify 3 loops: ABCDA, and BEFCB ABEFCDA. For each of these loops there is an associated equation, obtained by the loops law, but only two of those equations will be independent. In the case of the ABCDA loop, the loops law implies:
It is easy to verify the above equation, taking into account that , , and .
5.3. Loops method
In the circuits with resistors studied in Chapter 3, it was always possible to replace the resistors by an equivalent resistor and thus calculate the current supplied by the source and all the currents in the resistors.
In cases where there are several sources or when it is not possible to combine resistors (or capacitors) in series or in parallel, in order to obtain a single equivalent resistor (or capacitor) it is helpful to use the loops method. For example, in the circuit of Figure 5.4 none of the resistors is in series or in parallel with any other. Therefore, it is not possible to combine the resistors until to botain a single equivalent resistor.
We will use the circuit of Figure 5.4 to illustrate the loops method. When solving a problem, it will not be necessary to do a detailed analysis like the one below, but simply to use the rules given by the end of this section, in order to write the circuit matrices.
As shown in Figure 5.5 , one starts by identifying circuit 3 loops and each loop is assigned a loop current , and . Note that in Figure 5.5 , the loops are drawn with a rectangular shape, but are equivalent to the circuit in Figure 5.4 . It is convenient to select the same direction of rotation for all loop currents; in the case of Figure 5.5, we chose them all in the clockwise direction.
In resistances located between two neighboring loops, the current is the sum of the currents in the two loops. For example, in Figure 5.5 the current flowing through the resistor 5 Ω, the point A to C is (or the point C to A).
With this method the rule of the nodes is ensured in each node and just apply the rule of loops to each of the three loops to calculate the three currents. Potential differences between the various points of the circuit of Figure 5.5 , depending on the loop currents are the following (SI units):
replacing these values, the three equations of the loops are:
By grouping terms that depend on each of currents, you can write in matrix form the system:
The matrix system 5.5 was obtained by first calculating the potential differences in the circuit sections and applying the rule of the loops. However, you can write this system immediately looking at the circuit (Figure 5.5 ) and using the following rules:

Each line of the circuit matrix corresponds to one loop.

On the line The number in column is positive and the sum of all the resistances that is in loop .

The number on the line and column (with different than ) Is negative and absolute value equal to the sum of all the resistances that exist in the circuit segment that demarcates the border between the loops and .

each line the matrix with a column on the right side of the equation 5.5 is equal to the algebraic sum of all women that is in loop . In this algebraic sum are considered positive all sources in the sense arbitrated for current to flow from the negative electrode to the positive (potential increase) and negative all sources in the sense arbitrated for current to flow from the positive electrode to the negative (potential decrease).
Thus, the circuit is always symmetric matrix with positive elements on the diagonal and all other elements negative. In example 5.1 the rules set out above are used to directly write the matrix system of circuit equations.
The 3 loop currents are the solution of the system 5.5 , which can be obtained by any of the methods of solving systems of linear equations, for example, the rule of Cramer :
In this case, all the currents obtained are positive, which indicates that the sense of the loop currents coincides with the senses arbitrated in Figure 5.5 . In the circuit elements that do not belong to two loops, the actual current is equal current will of the respective network. Namely, the current flowing through the source is mA, the current on resistance is 3 Ω mA, and the current in resistor 4 Ω is mA (see Figure 5.5 ). The elements belonging to two loops have to combine the currents of the two loops for the actual current. For example, resistance of 5 Ω the current passes mA to the right and the current mA to the left; as such, the current in this resistor is to the right and intensely bad.
Example 5.1
In the circuit shown in the diagram compute: (a) the strength and direction of the current in resistance of 5.6 K ohms. (b) The potential difference in resistance of 3.3 K ohms. (c) power supplied or dissipated by each source.
Resolution. Let us start by choosing a consistent system of units so we can work with the numbers given, without having to write units in each equation. Expressing the values of resistances in kiloohms and the potential differences in volts, the current values will be in mA.
The circuit has 3 loops; however, one can reduce the number of loops to 2, because the resistance of 2.2 K ohms and 3.3 K ohms are in parallel and can be replaced by a single resistance: 2.2  3.3 = (2.2 × 3.3) / (2.2 + 3.3) = 1.32.
The equivalent circuit obtained with two loop currents is:
The corresponding matrix system in this circuit is:
Using Maxima , the system solution is:
(%o1) [ [ = −2.829, = −1.888] ]
and negative signals indicate that the two currents are in the opposite direction to the direction that has been arbitrated in the diagram.
(a) The resistance of 5.6 K ohms passes the current loop 1888, in order from A to B, and current loop 2829 in the direction B to A. Consequently, this current strength is 28291888 = 0941 mA from b to A. ( b ) the current resistance is 1.32 K ohms equal to the second current loop, 2,829 mA, C to B. Thus, the potential difference between C and b, which is also the potential difference resistance of 3.3 K ohms, is 1.32 × 3.73 = 2.829 V (greater potential than C B). (c) The current flowing through the source V 3 is equal to the first current loop, 1.888 mA; such that current flows from the positive electrode to the negative, the source of 3V dissipates a power of 1888 × 3 = 5664 mW. The source of 9 V, the current is equal to the second current loop, 2,829 mA; as this current passes from the negative electrode to the positive, the source provides a power of 2829 × 9 = 25.46 mW.5.4. Superposition Principle
In the example 5.1 , to each of the loop currents and It is separated into two portions, and The matrix equation 5.9 can be written as:
(5.10)
And, if current , , and are the two systems solutions:
(5:11)
It is guaranteed to and are the solutions of the equation 5.9 . These two systems of equations above correspond to two simpler circuit than the original circuit in Figure 5.6 , in each of the two new circuits of the sources being replaced by a wire with no resistance. These two new circuits are so simple, that can be solved without resorting to the method of the loops, as shown in the following example.
Example 5.2
Solve example 5.1 again, using the principle of superposition.
Resolution. Placing the source 9 V short circuit in Figure 5.6 , one obtains the following circuit:
current , and are the currents in the three resistors in units of mA. Note that since these are the actual currents and not the loop currents. The total resistance between the terminals of the source is:
As such, the current It is:
The potential difference across the parallel set ( 5.6 1.32) is
and the other two currents are:
Placing the source of 3V shortcircuited in Figure 5.6 , one obtains the following circuit:
The currents in the three resistors are now , and . Note that the current and They have opposite directions to the directions of and . The total resistance between the terminals of the source is:
and as such, the current It is:
The potential difference across the parallel set ( 1.2 5.6) is
and the other two currents are:
With these results and looking at the two circuit diagrams, can be calculated:
which are the same results obtained using the method of knitting. The rest of the resolution follows the same steps already made when the currents were calculated by the loop method.
5.5. Circuits with capacitors
The potential difference in a capacitor is directly proportional to the charge stored in its plate. If we connect a capacitor initially uncharged, between two points of a circuit, its initial potential difference is zero; it is as if, at that moment, a short circuit was made between the two points with a zeroresistance wire. In the following moments the potential difference increases as it enters the load capacitor; as the potential difference across the capacitor can not increase indefinitely, the load and the voltage listed reach final values.
When the load and the voltage across the capacitor to reach their final values, the current through the capacitor is zero and the condenser can then be considered as an open switch that does not let current. The increased load to the final value in the period in which the current decreases from the initial value to zero, is the response transient to the change produced by the source link.
The transient response is studied in the chapter on signal processing. In this chapter we consider only the initial and final values of electrical quantities in DC circuits. All capacitors in the circuit may be replaced by wire with zero resistance at the initial time0 = switch open and to calculate final values. The time required for the loads reach the final values is usually very short.
Example 5.3
A 1.2 μF capacitor, initially uncharged, is plugged to a battery with emf of 9 V and internal resistance of 230 Ω and a voltmeter with internal resistance of 3.2 kΩ is used to measure the voltage on the capacitor. (a) Find the initial and final currents in the battery. (b) Find the value of the final charge in the capacitor.
Resolution. The plugging of the capacitor to the battery can be represented by a switch that is initially open. The voltmeter must be connected in parallel to the condenser and, therefore, is represented by a resistance of 3.2 kΩ in parallel with the capacitor. The circuit diagram is then
(a) In the first moment, when closing the switch, the voltage of the capacitor and void because it is unloaded, equivalent to a wire with zero resistance; the equivalent diagram is the following
Note that all the current out of the source passes through this wire and no current flows through the voltmeter, because the wire resistance is zero. Another way to explain this result is that as the wire resistance is zero, the potential difference it is also null and as is in parallel with the voltmeter, the voltage of the voltmeter is zero (is in short circuit); the current in volimetro is it's like is zero, the current through the voltmeter is also zero. The current at the initial time is then equal to (SI units)
(%o2) 0.03913
When the capacitor is fully charged, the capacitor is equivalent to an open switch and the equivalent circuit is:
and final current is equal to
(%o3) 0.002624
(b) Since the condenser is connected in parallel with the voltmeter, the final potential difference between its plate is equal to the final potential difference in the resistance of 3.2 Ω, which is:
(%o4) 8.397
and the final charge on the capacitor is equal to
(%o5) 1.008×10^{−5}
that is, = 10,076 μC. Note that the results of Maxima commands show only 4 significant figures, but internally greater precision in the calculations and the values stored in the variables being used.
Example 5.4
In the circuit of example 5.1, if the resistance of 5.6 kΩ is replaced by a capacitor of 1.8 μF, find the final charge in that capacitor, and the sign of the charge in each of its sides.
Resolution. The circuit diagram is:
When the load reaches the final value, the capacitor acts as an open switch equivalent circuit is:
Thus, the circuit is also equivalent to a single source of 6V, in the same direction from the source 9, connected to a 2.52 K ohms resistance. The current is then counterclockwise direction and intensity equal to
This current allows to calculate the potential difference across the capacitor (equal to the potential difference between points A and B) and the load:
(%o6) 0.002381
(%i7) DV: 9  1320*I;
(%o7) 5.857
(%i8) Q: 1.8e6*DV;
(%o8) 1.054×10^{−5}
The final charge on the capacitor is then 10,543 μC and polarity is positive in equipment connected to point B and negative in the armature connected to the point A (the calculation DVwith the control %i7was made assuming a potential of B higher than the potential of A) .
Example 5.5
In the circuit shown in the diagram, determine the power dissipated in each resistance, and the energy stored in each capacitor.
Resolution. As the problem does not say for how long the capacitors have been connected to the sources, it is assumed that they have been connected long enough for the currents and charges to be in steady state. Thus, the capacitors are considered open switches and the equivalent circuit is as follows
The current in the 39 kΩ resistor is zero (no where to move) and the circuit has only a loop with total resistance 1.5 + 18 + 16 = 35.5 kΩ and current equal to
(%o9) 4.225×10^{−5}
From this current are calculated following all the power dissipated in the resistance and the energy stored in the capacitors.
Resistance of 39 K ohms, = 0, since the current is zero.
Resistance of 18 K ohms,
(%i10) P18: 18e3*I^2;
(%o10) 3.214×10^{−5}= 32,136 μW
Resistance of 16 K ohms,
(%i11) P16: 16e3*I^2;
(%o11) 2.857×10^{−5}= 28,566 μW
In the resistance of 1.5 K ohms,
(%i12) P1_5: 1.5e3*I^2;
(%o12) 2.678×10^{−5}= 2.678 μW
At 270 nF capacitor, the potential difference is the same as the resistance of 16 K ohms:
(%i13) DV: 16e3*I;
(%o13) 0.6761
(%i14) U270: 270e9*DV^2/2;
(%o14) 6.17×10^{−8}= 61,702 nJ

In the 180 nF capacitor, a possible route between the two points where it is connected past the source on the left, the resistance of 39 K ohms (with zero potential difference), the resistance of 1.5 K ohms and the second source. Therefore,
(%i15) DV: 1.5 + 1.5e3*I  1.5;
(%o15) 0.06338
(%i16) U180: 180e9*DV^2/2;
(%o16) 3.615×10^{−10}= 0.3615 nJ
Questions
(To check your answer, click on it.)
 Which of these physical principles associated with the law of nodes?
 Conservation of energy.
 Charge quantization.
 Conservation of charge.
 Conservation of momentum.
 Action and reaction.
 A capacitor in a DC circuit, which of the following magnitudes always have a final value null?
 Load.
 The potential difference.
 The current.
 The capacity.
 The stored energy.
 A constant voltage source was connected to a condenser and resistance 3, as shown in the diagram. What is the intensity of the current supplied by the source end?
 5 mA
 8 mA
 10 mA
 20 mA
 0
 if , and are the absolute values of currents flowing through the resistances , and
in the circuit of figure, which the equations is correct?
 + =
 + =
 + =
 =
 =
 Which of the following statements about the potential at points A, B and C, correct?
 > >
 > >
 > >
 > >
 > >
Problems
 In the circuit of Figure, determine which of the emf sources provide or absorb energy and calculate the power delivered or absorbed by each.
 Two identical cells, 9V, have been used differently, and therefore, one is more worn. The two cells bind in parallel to a resistance of 2.7 K ohms, as shown in the diagram. (a) Which of the two batteries is more worn? (b) Which of the two battery provides more power in the circuit?
(c) Calculate the current in the resistance of 2.7 K ohms.  If the two previous problem of the batteries were connected in series, not in parallel, which would provide them greater power in the circuit? That there may be inconvenient from a practical point of view?
 Determine the power dissipated in the circuit and each heater power provided by the emf Check that the power supplied by the emf is equal to the sum of all power dissipated in the resistances.
 In the circuit shown in the diagram, calculate: ( a ) The initial currents in resistors and capacitors. (b) The final charges the capacitors, indicating their polarities.
 (a) Determine the intensity and direction of the initial current in the condenser. (b) Determine the final charge on the capacitor and indicate polarity.
 In issue 4 , the resistance of 100 Ω is replaced by a capacitor of 39 nF, whose final energy stored in this condenser?
Answerss
Questions: 1. C 2. C 3. B 4. C. 5. D.
Problems
 The two sources provide power; emf 6 V provides 5.24 mW, and the emf of 5 V provides 3.93 mW.
 (a) What has internal resistance of 30 Ω. (b) What is less worn (with internal resistance of 20 Ω). (c) 3:32 mA.
 The two batteries provide the same power. The drawback is that when the battery more spend to fully discharge, the circuit will stop working, despite the other stack still load charged.
 At the resistance of 20 Ω, 45 μW. At the resistance of 100 Ω, 62.0 mW. In the resistance 150 Ω, 82.1 mW. In the resistance 60 Ω, 105.8 mW. At the resistance of 80 Ω, 151.4 mW. The emf provides 401.4 mW.
 Using subscripts equal to the resistance value or capacity ( a ) = 0, = 7.5 mA, = = 40 mA = 32.5 mA. (b) = 0, = = 7.1 mA, = 87.4 nC (positive in the low and negative reinforcement on top) NC = 597.4 (positive to the right of the armature and negative from the left).
 (a) 4.78 mA upwards. (b) 2:44 μC (negative in plate up and positive on the bottom).
 236.5 nJ.
Loads that reach a node have to go through any of the possible paths connected to that node; this has nothing to do with the energies of the charges.
(Click)