Abstract
An electrical circuit, or electrical network, consists of a set of physical devices interconnected so as to allow the flow of the electrical current. The devices used in electrical circuits are called electrical components or circuit elements. Note that the term circuit highlights a key feature of an electrical network, consisting in the presence of one or more closed paths through which the current can circulate.
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Notes
- 1.
Nowadays it is customary to use the term electronic for all electrical components even if electronics itself was founded in the early 1900s with the invention, by A. Fleming and L. De Forest, of vacuum valves, the first devices able to control the motion of electrons.
- 2.
The electromagnetic field propagates in electrical circuits with a speed of the order of light speed in vacuum c. Denoting with \(\nu \) the frequency of the signal, its wavelength is \(\lambda \simeq c/\nu \). To get an idea of the order of magnitude of lengths, consider that for \(\nu =1\) kHz, \(\lambda =3.010^5\) m, for \(\nu =1\) MHz, \(\lambda =3.010^2\) m and for \(\nu =1\) GHz, \(\lambda =0.3\) m. These lengths are larger than typical dimensions of components \({\simeq }10^{-2}\div 10^{-3}\) m and of circuits \({\simeq }10^{-1}\) m.
- 3.
A simple model of the electrical conduction due to Drude provides the behavior of ohmic conductors. The model is based on the assumption that, inside the conductor, the electrons (point-like particles with electrical charge e) are subject to two forces. The first is due to the external (constant) electric field (\(F_D = eE\)). The second is a viscous force proportional to the electron speed and takes into account the diffusion effects suffered by the electrons in the solid ( \(F_V=-\gamma v\)). In one spatial dimension we can write \(m_e\dot{v}=eE-\gamma v\). In stationary conditions the electron drift velocity (\(v_D\)) is \( v_D=eE/\gamma \), proportional to the electric field. Since the current density J is given by \(J=nev_D\), where n is the electron particle density, we have \(J=ne^2E/\gamma \) . If A is the conductor cross sectional area and l its length, the electric field is \(E=V/l\) and the current is \(I=JA=(ne^2/\gamma ) (A/l) V\). The first Ohm’s law is thus recovered with the resistance \(R=(\gamma /ne^2)(l/A)\). Furthermore, we note that the quantity \(\gamma /ne^2\) represents the material resistivity \(\rho \) as defined by the second Ohm’s law \(R=\rho l/A\).
- 4.
The equality \(L_{jk} = L_{kj}\) is recovered rather easily when the coefficients \(L_{jk}\) are derived using the vector potential \(\varvec{A}\). Details can be found in general physics textbooks, as for example The Physics of Feynmann Vol. II-17-11 [1].
- 5.
In this notation, \(L_{jj}\) is the self-inductance of the jth inductor.
- 6.
The correspondence between the notation used in this example and the one used previously is as follows: \(L_1=L_{11}, L_2=L_{22}\) and \(M=L_{12}=L_{21}\).
- 7.
- 8.
Note that parasitic effects in real component depend on their geometry and extension. Therefore, a representation in terms of discrete elements is always an incomplete approximation of the component behavior that could be better described in terms of distributed constants.
- 9.
The temperature is an influence variable for the value of R. As we will see in detail in the chapter dealing with the handling of measurement uncertainties, the influence variables are physical quantities that, although not directly involved in the definition of the physical variable of primary interest, in this case the resistance, can have an influence on their value.
- 10.
Data taken from: “https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity.” This site contains references to many texts and other web sites on resistivity values.
- 11.
An electrolytic capacitor incorrectly connected, with the wrong polarity, is permanently damaged and can create a dangerous situation with the emission of toxic fumes and with the possibility of component explosion.
- 12.
The practical implementation of such a device requires familiarity with working principles of operational amplifiers and cannot be discussed here.
- 13.
Only for a few geometries we can easily calculate the inductance. Among these: the toroidal winding and, neglecting edge effects, the winding on a cylinder.
- 14.
Here it is supposed to use the inductor with a sinusoidal generator of frequency \(\nu =\frac{\omega }{2\pi }\). Periodic currents will be treated extensively in Chap. 5.
- 15.
More detailed discussion on the skin effect can be found in electrodynamics textbook such as, for example, [4].
References
R.P. Feynmann, R.B. Leighton, M. Sands, The Feynmann Lectures on Physics, vol. II (Dover Publications, New York, 1967)
Munir H. Nayfeh, Morton K. Brussel, Classical Electricity and Magnetism, 2nd edn. (Dover Publications, Inc., Mineols New York, 2015)
K.H. Wolfgang, Panofsky and Melba Phillips, 2nd edn. (Melba Phillips, Classical Electricity and Magnetism 2005)
J.D. Jackson, Classical Electrodynamics, 3rd edn. (Academic Press, New York, 1998)
J.D. Romano, R.H. Price, Am. J. Phys 64, 1150 (1996)
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Problems
Problems
Problem 1
A \(220\,\Omega \) resistor can dissipate up to \(0.5\,W\) without damage. Compute the maximum voltage and the maximum current that it can withstand. [A. \(V_\mathrm{max}=10.5\,\mathrm{V},\, I_\mathrm{max}=47.7\,\mathrm{mA}\).]
Problem 2
Calculate the amount of energy that can be accumulated in a capacitor of 12 nF whose dielectric can sustain a maximum voltage of 50 V. [A. \(1.5 \times 10^{-5}\,\mathrm{J}\).]
Problem 3
Calculate the amount of energy that can be accumulated in an inductor of 15 mH that can sustain a maximum current of 0.5 A. [A. \(1.87 \times 10^{-3}\,\mathrm{J}\).]
Problem 4
The battery of your car has 0.4 MJ of stored energy, 12 V of open circuit voltage and an internal resistance of \(1.0\,\Omega \). If you forget the lights on and the lamps have an equivalent resistance of 5 \(\Omega \), after how long will the battery be fully discharged? Use the simplifying assumption that the internal resistance of the battery remains unchanged until the discharge completes. [A. 4 h38 m.]
Problem 5
Calculate the resistance of a 1000 W electric heater, recalling that the electricity network delivers an effective voltage of 220 V. [A. \(48.4\,\Omega \).]
Problem 6
A voltage generator with output V connects to the series of two resistors \( R_1\) and \(R_2\) forming a voltage divider. Compute the voltage drop across each resistor and their ratio. [A. \(\varDelta V_1=V R_1/(R_1+R_2), \varDelta V_2=V R_2/(R_1+R_2), \varDelta V_1/\varDelta V_2=R_1/R_2\).]
Problem 7
N resistors with resistance values \(R_n\) are connected in series to form a chain; calculate the voltage drop \(\varDelta V_n\) across each resistor \(R_n\) when voltage V is applied to the chain. [A.\(\, \varDelta V_n = V R_n/\sum _iR_i\).]
Problem 8
A current generator with output I connects to the parallel of two resistors \( R_1\) and \(R_2\) that form a current divider. Compute the current value in each resistor and their ratio. [A. \(I_1 = IR_2/(R_1 + R_2), I_2 = IR_1/(R_1 + R_2), I_1/I_2 = R_2/R_1\).]
Problem 9
N resistors with resistance values \(R_n\) are connected in parallel. Compute the current value in the generic n-th resistor as a function of the total current I supplied by the generator powering the circuit. [A.\( I_n = I/R_n/\sum _i(1/R_i)\).]
Problem 10
Compute the voltage \(V_{AB}\) in the circuit shown in the figure using the expressions deduced for the voltage divider in problem 6. Assume \(V=100\,\mathrm{V},\, R_1=100\,\Omega , R_2=R=200\,\Omega \). [A. \(V_{AB}=50\,\mathrm{V}\).]
Problem 11
Assume to have a 7 A current generator and three resistors of values \(0.25, 0.5, 1\,\Omega \) respectively. How would you arrange them to build a current generator with output equal to 1 A? [A. Connect the three resistor in parallel and take the output in series with the \(1\,\Omega \) resistor.]
Problem 12
Compute the power dissipated by each of the three resistors in the previous problem. [A. \(P_{0.25\Omega } = 4\mathrm{W}; P_{0.5\varOmega } = 2\mathrm{W}; P_{1\varOmega } = 1\,\)W.]
Problem 13
Compute the current in the resistor \(R_3\) in the circuit of the figure using the expression deduced for the current divider in problem 8 assuming \(R_1=100\,\Omega ,R_2=R_3=50\,\Omega \). [A. \(I_{R_3}=2.5\,\mathrm{A}\).]
Problem 14
Compute the power delivered by the generator and the power dissipated in the resistor \(R_1\) of the circuit in the figure. [A. \(W_g = V^2/R_{eq}, W_{R_1} = V^2R_1/R_{eq}^2\) con \( R_{eq}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3+R_4}}\).]
Problem 15
Compute the power delivered by the current generator and the power dissipated in the resistor \(R_2\) of the circuit in the figure. [A. \(W_g = I^2(R_1 +R_2)(R_3 +R_4)/(R_1 +R_2 +R_3 +R_4), W_{R_2} = I^2R_2(R_3 +R_4)^2/(R_1 +R_2 +R_3 +R_4)^2)\).]
Problem 16
Compute the resistance between the two points A and B and the resistance between the two points C and D of the circuit in the Figure. [A. \(R_{AB}= R_1+R_3+\frac{1}{\frac{1}{R_2}+\frac{1}{R_6}+\frac{1}{R_4+R_5}}, R_{CD}=\frac{1}{\frac{1}{R_5}+\frac{1}{R_4+\frac{1}{\frac{1}{R_2}+\frac{1}{R_6}}}}\).]
Problem 17
Two capacitors with capacitance \(C_1\) and \(C_2\) are connected in series and the series is connected to a voltage generator of output V. Compute the charge accumulated on each of them. [A. \(Q=V C_1 C_2/(C_1+C_2\).]
Problem 18
Two capacitors with capacitance \(C_1\) and \(C_2\) are connected in series and the series is connected to a voltage generator of output V. Compute the voltage drop across each capacitor and their ratio. [A. \(\varDelta V_1=V C_2/(C_1+C_2), \varDelta V_2=V C_1/(C_1+C_2), \varDelta V_1/\varDelta V_2=C_2/C_1\).]
Problem 19
Two inductors with inductance \(L_1\) and \(L_2\) are connected in series and the series is connected to a voltage generator of output v(t). Neglecting mutual inductance, compute the voltage drop across each inductor and their ratio. [A. \(\varDelta V_1=v(t) L_1/(L_1+L_2), \varDelta V_2=v(t) L_2/(L_1+L_2), \varDelta V_1/\varDelta V_2=L_1/L_2\).]
Problem 20
* The second Ohm’s law holds for resistors with a constant cross section. Consider how to solve the problem of computing the resistance R of a resistor having the form of a truncated cone with length l and a cross section radius ranging from \(r_1\) to \(r_2\). Assume that the resistor material is homogeneous with resistivity \(\rho \). Many textbooks give the answer: \(R = \rho l/\pi r_1r_2\); recover the procedure leading to this answer and show that it is wrong since it violates charge conservation. The full response can be found in Ref. [5].
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Bartiromo, R., De Vincenzi, M. (2016). The Electrical Circuit and Its Components. In: Electrical Measurements in the Laboratory Practice. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-319-31102-9_1
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