G. Las Reconstrucciones
1.5.5. La Musealización del Patrimonio Andaluz
Although the terms ‘series’ and ‘parallel’ have already been used in the previous sections on the assumption that readers know what they mean, they should be defined formally in order to remove any room for confusion. Two or more circuit elements are said to be connected in series if their currents are the same (in view of their reference directions). In order for two elements to be in series, they must have a common (simple) node shared by them exclusively so that any current leaving one element enters the other. Two or more circuit elements are said to be connected in parallel if their voltages are the same (in view of their reference polarities). In order for two elements to be in parallel, they must have a common pair of nodes.
When mentioning ‘series’ or ‘parallel’, usually it is only the physical connection that is of interest, without minding the reference direction of the current or the reference polarity of the voltage since they are arbitrary.
Especially for series and parallel connections of two resistors R1and R2, the following formulas are used to obtain their equivalent resistances, which will be derived in Section 2.1:
1. The equivalent resistance of two resistors in series is
Req;s¼ R1þ R2 ð1:33aÞ
2. The equivalent resistance of two resistors in parallel is Req;p¼ R1jjR2¼ 1
1=R1þ 1=R2
¼ R1R2
R1þ R2
ð1:33bÞ
Problems
1.1 Resistance and Resistivity
(a) Noting that the resistivity of copper is r¼ 1:68 108 m, find the resistance of a copper wire that is 1 m long and 1 mm in diameter.
(b) Find the inductance of an inductor that consists of a coil of insulated wire wound 100 turns around a core having the permeability of ¼ r0¼ 1000 4 107¼ 1:256103H=m, a diameter of 5 mm, and an average length of 5 cm.
Figure 1.21.5 Current-to-voltage source transformation using series duplication of the I-source
(c) Find the capacitance of a capacitor consisting of two parallel square metal plates, each 1 cm 1 cm in size, between which a 1 mm-thick ceramic dielectric with a dielectric constant (permittivity) of "¼ "r"0¼ 1000 8:854 1012¼ 8:854 109F= m is inserted.
1.2 Permissible Voltage/Current under Power Rating
(a) Suppose there is an incandescent electric lamp whose rated power and voltage are 100 W and 100 V, respectively. What is the magnitude of the current flowing through it under the rated voltage? Also find its resistance.
(b) Find the voltage and current that are allowed to be applied continuously for a 1 k resistor with the power rating of 10 W.
1.3 Load Resistors in Series and Parallel
(a) Find the powers of the two resistors 1 and 2 in series with a 12 V voltage source in the circuit of Figure P1.3(a).
(b) Find the powers of the two resistors 1 and 2 in parallel with a 12 V voltage source in the circuit of Figure P1.3(b).
(c) Circle the correct word in each set of parentheses in the following statement: Combining the results obtained in (a) and (b), it is conjectured that less power is dissipated by (smaller, larger) resistance in a series connection and (larger, smaller) resistance in a parallel connection.
1.4 Sign of Power and Passive Sign Convention
Consider the circuit of Figure P1.4 in which the mesh current i circulating through the mesh is found to beð12 3Þ=3 ¼ 3 A in its (clockwise) reference direction.
(a) Find the current iR1 through the resistor R1with its sign in view of its reference direction depicted in the circuit diagram. Find also the voltage vR1across the resistor R1with its sign in view of its reference polarity.
(b) Find the power of the 3 resistor, which should be nonnegative because a resistor can supply no energy even instantly. What is wrong?
Figure P1.3
Figure P1.4 26 Chapter 1 Basic Concepts on Electric Circuits
(c) Find the powers of the two voltage sources Vs1and Vs2and tell whether each of them delivers or absorbs the power.
(d) Find the total power of the two voltage sources and the 3 resistor. What does the result imply?
1.5 Ohm’s Law, KVL, and KCL for an R–2R Ladder Network Consider the R–2R ladder circuit of Figure P1.5.
(a) To find the output voltage Vofirst, assume it to be known, and then express all the branch currents and node voltages including v1in terms of Vo, starting from the branch farthest away from the source, via the repeated use of Ohm’s law, KCL, and KVL. Lastly, use v1¼ 12 V to obtain Vo. (b) To find the input current i1first, apply the series and parallel combination formulas for resistors to find the overall equivalent resistance of the ladder network seen from the input terminals 1–0.
Then find the input current i1, the node voltage v2, the branch currents i2and i3, the node voltage v3, the branch currents i4and i5, and finally the output voltage v4¼ Vo.
1.6 KVL/KCL, Tellegen’s Theorem, Node Voltages, Mesh Currents, and Branch Voltages/Currents Consider the circuit of Figure P1.6(a) in which the node 0 is grounded so that its node voltage is zero, that is v0¼ 0. Note that the grounded node is called the reference node.
(a) Apply KVL around mesh 1 to write a KVL equation in the branch voltages va, vb, and vc. Referring to Figure P1.6(b), note that the branch voltages va, vb, and vccan be expressed in terms of the node voltages v1and v2as follows:
va¼ v1 v0¼ v1 0 ¼ v1; vb¼ v1 v2; and vc¼ v2 v0¼ v2 0 ¼ v2 ðP1:6:1Þ Substitute these expressions into the KVL equation for mesh 1 to check whether it is satisfied.
Note. This implies that just expressing branch voltages in terms of node voltages is based on KVL and thus it corresponds to applying KVL implicitly.
Figure P1.5 R–2R ladder network
Figure P1.6
(b) Apply KCL to node 2 to write a KCL equation in the branch currents ib, ic, and id. Referring to Figure P1.6(c), note that the branch currents ib, ic, and idcan be expressed in terms of the mesh currents i1and i2as follows:
ib¼ i1; ic¼ i1 i2; id¼ i2 ðP1:6:2Þ Substitute these expressions into the KCL equation for node 2 to check whether it is satisfied.
Note. This implies that just expressing branch currents in terms of mesh currents is based on KCL and thus it corresponds to applying KCL implicitly.
(c) The total power of all the elements in the circuit can be written as
p¼ vaiaþ vbibþ vcicþ vdidþ veie ðP1:6:3Þ Verify that substituting the expressions of the branch voltages in terms of the node voltages into this equation yields
p¼ v1ðiaþ ibÞ þ v2ðicþ id ibÞ þ v3ðie idÞ ðP1:6:4Þ which turns out to be zero with the KCL equation at every node. This implies Tellegen’s theorem, described by the following equation:
Xb
k¼1
pk¼Xb
k¼1
vkik¼ 0; b: the number of branches ðP1:6:5Þ
1.7 KVL and KCL for a Circuit
Consider the circuit of Figure P1.7 in which the node 0 is grounded so that its node voltage is zero, that is v0¼ 0.
(a) After removing every source by short-circuiting the voltage source(s) and open-circuiting the current source(s), find b, the number of branches, and n, the number of nodes. Referring to Section 1.4.4, find the number of independent KCL and KVL equations.
(b) Unlike the current through the voltage source Vs1, the currents through the resistors R1, R2, and R3can be expressed in terms of the node voltages v1¼ Vs1, v2, and v3based on Ohm’s law as
iR1 ¼v1 v2
R1
¼Vs1 v2
R1
; iR2 ¼v2 v3
R2
; and iR3 ¼v0 v3
R3
¼ v3
R3
ðP1:7:1Þ Figure P1.7
28 Chapter 1 Basic Concepts on Electric Circuits
Find the appropriate nodes (including any supernode) to which KCL can be applied. Write the KCL equations in the branch currents iR1; iR2, and iR3at those nodes. Substitute the above VCRs (voltage–current relationships) of the branch elements into the independent KCL equation(s) to find the node equations in the node voltages v2and v3. Are there as many independent KCL equations as estimated in (a)?
(c) Unlike the voltage across the current source Is1, the voltages across the resistors R1, R2, and R3
can be expressed in terms of the mesh currents i1and i2based on Ohm’s law as
vR1¼ R1iR1¼ R1i1; vR2¼ R2iR2 ¼ R2i2; and vR3 ¼ R3iR3 ¼ R3i2 ðP1:7:2Þ Find the appropriate mesh(es) or loop(s) around which KVL can be applied. Write the KVL equation(s) in the branch voltages vR1; vR2, and vR3 around the mesh(es) and/or loop(s). Is there any loop that can be called a supermesh? Substitute the above VCRs (voltage–current relationships) of the branch elements into the KVL equation(s) to find the mesh equations in i1 and i2. Are there as many independent KVL equations as estimated in (a)? Is the number of independent KVL equations the same as that of the unknown mesh currents, thus enabling the KVL equation(s) to be solved for i1and i2? If not, write another equation i2 i1¼ Is1, which is presented by the current source Is1 involved in a supermesh, i.e.
shared by the two meshes and playing the role of matchmaker to relate the two mesh currents i1and i2.
1.8 KVL and KCL for a Circuit Consider the circuit of Figure P1.8.
(a) After removing every source by short-circuiting the voltage source(s) and open-circuiting the current source(s), find the number of branches, b, and the number of nodes, n. Referring to Section 1.4.4, find the number of independent KCL and KVL equations.
(b) Apply KCL to the appropriate nodes (possibly including any supernode) to write as many independent KCL equations as possible in the three unknown node voltages v2, v3, and v4, where v1¼ Vs1is already known. Is there the same number of KCL equations as estimated in (a)? Is the number of independent KCL equations the same as that of the unknown node voltages, thus enabling the KCL equation(s) to be solved for v2, v3, and v4? If not, use another equation v2 v3¼ Vs2, or v2¼ v3þ Vs2, which is presented by the voltage source Vs2
involved in a supernode, i.e. shared by the two nodes and playing the role of matchmaker to relate the two node voltages v1and v2.
(c) Apply KVL to the appropriate meshes or loops (possibly including any supermesh) to write as many independent KVL equations in the three unknown mesh currents i1, i2, and i3, where
Figure P1.8
i4¼ Is2is already known. Is this the same number of KVL equations as estimated in (a)? Is the number of independent KCL equations the same as that of the unknown mesh currents, thus enabling the KVL equation(s) to be solved for i1, i2, and i3? If not, use another equation i3 i2¼ Is1, or i3¼ i2þ Is1, which is presented by the current source Is1 involved in a supermesh, i.e. shared by the two meshes and playing the role of matchmaker to relate i2
and i3.
(d) With R1¼ 1 , R2¼ 2 , R3¼ 3 , R4¼ 4 , Vs1¼ 12 V, Vs2¼ 6 V, Is1¼ 20 A, and Is2¼ 23 A, solve the set of KCL equations for v2, v3, and v4and use the solution to find iR1, iR2, iR3, and iR4. Also solve the set of KVL equations for i1, i2, and i3and use the solution to find iR1, iR2, iR3, and iR4. Do the two solutions agree with each other?
1.9 KCL, KVL, and the Source Combination Consider the circuit of Figure P1.9.
(a) Applying the rules of combining the voltage source and current source described in Figures 1.18.7 and 1.18.8, remove the sources that are dispensable for finding the current iR2through R2. (b) Apply KCL to node 3 to write a KCL equation in iR1and iR2, substitute the expressions of the branch currents iR1and iR2in terms of the node voltage v3into the KCL equation, and solve it for v3. Use the result to find the current iR2 through the resistor R2.
(c) Apply KVL to mesh 2 to write a KVL equation in vR1and vR2, substitute the expressions of the branch voltages vR1 and vR2 in terms of the mesh currents i2and i3into the KVL equation, substitute i3¼ Is2, and solve it for i2. Use the result to find the current iR2through the resistor R2.
1.10 Source Transformation and Equivalent Circuit for Parallel Voltage Sources with a Resistor in Series Consider the circuit of Figure P1.10(a) in which two voltage sources each having a resistor in series are connected in parallel and applied to a load resistor RL.
(a) Apply KCL to node 1 to write a KCL equation in iR1, iR2, and iRL, substitute the expressions of the branch currents iR1, iR2, and iRLin terms of the node voltage v1into the KCL equation, and solve it for v1. Find the current iRLthrough RL.
(b) Referring to the source transformation introduced in Section 1.5.2, transform each of the two voltage sources (with a resistor in series) into a current source (with a resistor in parallel), as shown in Figure P1.10(b). Then combine the two parallel current sources into an equivalent one (referring to Section 1.5.1) and the two parallel resistors into an equivalent one (referring to Section 1.6) so that an equivalent of the source part of the original circuit seen from the terminals 1–0 is obtained, as shown in Figure P1.10(c). Express the values of the equivalent voltage source Vsand the equivalent resistor R in terms of Vs1, Vs2, R1, and R2. Using this equivalent circuit, find the current iRL through RL.
Figure P1.9 30 Chapter 1 Basic Concepts on Electric Circuits
1.11 Simplification of a Circuit by Source Transformation
To see how source transformation can be used to simplify circuits, the source transformation technique will be applied successively for the circuits that were obtained in Figures 1.21.2 to 1.21.5.
(a) Figure P1.11.1(a) shows the circuit in Figure 1.21.2(b). The 42A source with the 1 resistor in parallel can be transformed into a 42V source in series with the 1 resistor, combined with the 9V source to make a 51V source, as in Figure P1.11.1(b), and transformed into a 51A source with the 1 resistor in parallel, as in Figure P1.11.1(c). Lastly, the two resistors (of 1 and 2) and the three current sources (of 51A, 21A, and 3A) in parallel can be transformed as a whole into a ( )V-source in series with a ( )-resistor, as depicted in Figure P1.11.1(d).
(b) Figure P1.11.2(a) shows the circuit in Fig. 1.21.3(b). The 9V source with the 1 resistor in series can be transformed into a 9A source in parallel with the 1 resistor (Figure P1.11.2(b)) and combined with the 3A source and the 2 resistor in parallel to make a ( )A-source in parallel with a ( ) resistor, as in Fig. P1.11.2(c). Lastly, the 6A source with the (2/3)
resistor in parallel can be transformed into a 4 V source in series with the (2/3) resistor and then combined with the 42V source in series, as depicted in Figure P1.11.2(d).
(c) Figure P1.11.3(a) shows the circuit in Figure 1.21.4(b). The 9V source with the 1 resistor in series can be transformed into a 9A source in parallel with the 1 resistor, the three voltage sources of 42V, 9V, and 12Vare combined into a 63V source, as in Fig. P1.11.3(b), and then the two resistors (1 and 2) in parallel are combined to make one resistor of (2/3), as in Figure P1.11.3(c). Lastly, the 9A source in parallel with the (2/3) resistor can be transformed into a ( )V-source in series with the ( ) resistor and then combined with the 63V source in series as depicted in Figure P1.11.3(d).
Figure P1.10 Simplification of the circuit using the source transformation
Figure P1.11.1
Figure P1.11.2
Figure P1.11.3
Figure P1.11.4 32 Chapter 1 Basic Concepts on Electric Circuits
(d) Figure P1.11.4(a) shows the circuit in Figure 1.21.5(b). The 9V source with the 1 resistor in series can be transformed into a 9A source in parallel with the 1 resistor, and the 6V source with the 2 resistor in series into a 3A source in parallel with the 2 resistor (Figure P1.11.4(b)). Then the two current sources of 9A and 3A in parallel and two resistors of 1
and 2 in parallel can be combined to make a ( )A-source in parallel with a ( )
-resistor, as in Figure P1.11.4(c). Lastly, the 6A source with the (2/3) resistor in parallel can be transformed into a 4V source in series with the (2/3) resistor and then combined with the 42V source in series, as depicted in Figure P1.11.4(d).
(e) Simplify the circuit in Figure 1.21.1 by starting from the transformation of the 9V source in series with the 1 resistor into a 9A source in parallel with the 1 resistor.
(f) Let the objective be to find the current through the 4 resistor. Among the equivalent circuits obtained in (a), (b), (c), (d), and (e), find one that does not serve the purpose and explain the reason why it does not.
Note. This problem implies that it is better not to touch the target element (whose voltage or current we are interested in) when using the equivalence to simplify a circuit.