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Electric Charge Current and Voltage, ways on how to solve them
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Faculty, College of Marine Engineering University of Cebu Lapu-lapu and Mandaue Cebu City, Philippines [email protected]
Fig. 1. Bohr Atomic Model (© 2012 Encyclopedia Britannica Inc.)
A matter defines as an object that occupies space and with finite mass. It is made up of countless dense particles moving in a finite space. A matter breaks into a compound or an element. A compound is a combination of two or more set of elements. A molecule is the smallest representation of a compound. Each element consist of a unique set of particles, known as an Atom. An atom is a fundamental building block of matter. It consist of three elementary particles: the neutron n, the proton p+^ and the electron e−.In the atomic lattice, neutron and proton makes up the nucleus, while electrons orbits around the nucleus in a fixed orbit (see F ig. 1 ). Proton and electron are positively and negatively charge particles. The number of p+^ and e−^ in an atom are equal so that it remains at a neutral state. A deviation in the number of e− results to a charged atom, also known as an Ion. When an atom losses an e−, it results to a positively charge atom, known as Cation. In contrast, when atom gains an e−, it results to a negatively charge atom, known as Anion. An electronic model of an atom maps the e−^ to its shell or orbit (see F ig. 2 ). Each orbit or shell holds a finite number of e−. The e−^ in a given shell is N = 2n^2 where N is maximum number of e−^ in the orbit and n is nth^ shell or orbit. A valence electron lies at the outermost shell or orbit of an atom. It plays an important role to electricity. When we apply sufficient potential on an atom, the valence e−^ moves
Fig. 2. Electronic Model of an Atom (©Byjus.com)
to the neigbhoring atom causing electron flow or electricity. We call this process as Ionization. Each orbiting e−^ associates with a discrete energy level. The energy state is directly proportional to its distance from the nucleus. The e−^ closer to the nucleus has lower energy state than at the outermost orbit (valence e−). Also, the e− that moves out from the atomic structure experiences higher energy state that e−^ in the atomic structure. Interestingly, we see a parallelism between our solar system and the atomic structure. In physics, the gravitational force, F = Gm d^12 m 2 , is greater at object nearest to the sun than those farthest from it. At the atomic level, e−^ nearest to the nucleus requires more energy to move from one orbit to another as to valence e−.
II. ELECTRIC CHARGE An electric charge is a fundamental quantity of an electric circuit and measured in coulombs (C). A charge e−^ is negative and equal to a magnitude of 1. 602 × 10 −^19 C while p+^ carries a positive charge of same magnitude. A 1 C of charge is equal to 6. 24 × 1018 e−. Experiments find out that only integral multiples of electronic charge is observable in nature. These charges can niether be created nor destroyed, only transfered from one atom to another (Law of Conservation of Charge). Also, unlike charges repels, and like charges attracts. Let us have an example. Example 1: A metallic sphere and a negatively charged rod has a charge of +4 nC and − 6 nC, respectively. When the rod touches the sphere, 8. 2 × 109
electrons are transferred. What are the charges of the sphere and the rod now? We start with modelling an equation that mathematically express the problem. Let QS and QR the charges experience at the sphere and the rod after the rod touches the sphere, respectively. Hence stated in the problem, the rod loses charges after it touches the sphere, while the sphere gains the lose charges from the rod. We derive
QS = QS 0 + Q∆ QR = QR 0 − Q∆
where QS 0 , QR 0 is the original charge, and Q∆is an amount of lose or gain charge by the sphere and the rod.
Q∆ = 8. 2 × 109 e−^ ×
From (1), we get the values of QS and QR.
QS = 4 nC + (− 1. 314 nC) = 2. 686 nC QR = − 6 nC − (− 1 .314) nC = − 4. 686 nC
Thus, the rod and sphere have − 4. 686 nC and 2. 686 nC, respectively. Let us have another one. Example 2: A positively charge dielectric has a charge of 2 C. If a cloud of 12. 5 × 1018 electrons are added into it, what will be the net charge? Let Qnet is the resulting charge of the dielectric.
Qnet = Qd + Qe (4)
where Qd and Qe is the charge of the dielectric and the cloud of electrons, respectively.
Qe = 12. 5 × 1018 e−^ ×
Qnet = 2 C + (− 2 C) = 0 C
The Qnet is equal to 0 C.
Previously, we discussed that a suffiecient potential moves the valence e−^ to the neigbhoring atom which cause electron to flow. Electric current is the time rate of electron flow, measured in Ampere (A). Mathematically, it is the relationship between current i, charge q and time t.
i = dq dt
where 1 A is equal to 1 Cs. We obtain charge Q by integrating the equation (6) between t 0 and t.
q =
∫ (^) t
t 0
idt (7)
Electric current is not a single-valued function since charge varies with time in different way. If the value of current does not change over a period of time (see F ig. 3 ), it is a direct current (dc). We often use direct current to power smartphones,
Fig. 3. Direct Current
Fig. 4. Alternating Current
laptop and other electronics devices. In this text, we use I to denote direct current. If a current varies sinusoidally with time (see F ig. 4 ), it is an alternating current (ac). It is use in households to run elecric machines, air-conditioning units and other electric appliances. We represent a time-varying current as i. Hence electric current is the flow of electron, we associate current with its direction of flow. Electron and conventional flow are two convention to define the flow of charges in an electric circuit. Electron flow take the actual flow of charges from lower (negative) to higher (positive) potential. Conventional flow is widely used in analysis of electric circuit. In conventional flow, charges flow in reverse. It assumes that charges flow from higher to lower potential. In this text, we follow conventional flow in our analysis. Let us have some example. Example 3: A cloud of 12. 5 × 1018 electrons move past a given point in every 2 sec. How much is the intensity of electron flow?
At t = 10 ms,
q − q 1 =
( (^) q 2 −^ q 1 t 2 − t 1
t − t 1
q − 80 mC =
( (^0) mC − 80 mC 12 ms − 8 ms
t − 8 ms
q = − 20 t + 168
i =
d dt
(− 20 t + 240) = − 20 A
Example 9: Find the velocity of charge leading to 1 A current which flows in a copper of conductor of cross-section 1 cm^2 and length of 10 km. Free electron density of copper is 8. 5 × 1028 e−^ per m^3. How long will it take the electric charge to travel from one end of the conductor to the other?
e− m^3
m^2
10000 m
t
1 A =
t t = 1. 3622 × 1010 s
velocity =
distance time
10000 m
velocity = 7. 3412 × 10 −^7
m s (18)
VI. PROBLEM SET
A current of 3. 2 A flows through a conductor. Calculate how much charge passes through any cross-section of the conductor in 20 s. ANS. 64 C
The current through an element is shown in F ig. 6. Determine the total charge that passed through the element at: (a) t = 1 s, (a) t = 3 s, and (a) t = 5 s. ANS. (a) 10 C, (b) 22. 50 C, (c) 30 C
A lightning bolt with 8 kA strikes an object for 15 μs. How much charge is deposited on the object? ANS. 120 mC
A rechargeable flashlight battery is capable of delivering 85 mA for about 12 hr. How much charge can it release at that rate? If its terminals voltage is
Fig. 6. Problem Set 2
Fig. 7. Problem Set 5
ANS. 15 mC
Given that the number of charges measured at t 1 is n 1 = 2 × 1019 , and the number of charges measured at t 2 is, n 2 = 5. 75 × 1019. Find the current flow if the time interval between t 1 and t 2 is 2 s, with the assumption that the current, i(t) is a linear function of charges, q(t). ANS. 3 A
A depth of 0. 15 mm nickel is required to be coated on a metal surface by electroplating. The nickel density r = 8. 8 × (^103) mkg 3 , the electrochemical equivalent k = 0. 304 × 10 −^6 kgC (which is the weight of the nickel produced by electrolysis during the flow of a quantity of electricity equal to 1 C), and the current per square meter r = 35 (^) mA 2. How long (in hours) does it take to achieve the desired coating depth? ANS. 34. 46 hr
If an alternating current of i(t) = 2 sin( 32 πt) flows from a voltage source of v(t) = 5 V , find the energy used in 1 s. ANS. 0. 41 J
The charge entering the positive terminal of an element is q = 10 sin 4πt mC while the voltage across the element (plus to minus) is v = 2 cos 4πt V. (a) Calculate the energy delivered to the element between 0 and 0. 6 s. (b) Find the current for t = 0. 6 s. ANS. (a)78. 34 mJ and (b)124. 58 mA
Charles Alexander and Matthew Sadiku. 2008. Fundamentals of Electric Circuits (4th. ed.). McGraw Hill Higher Education.
Stan Gibilisco. 2006. Teach Yourself Electricity and Electronics, Fourth Edition (4th. ed.). McGraw-Hill, Inc., USA.