TJC Unit 15 Currents

Student’s Copy Unit 15: Currents Learning Outcomes Candidates should be able to: (a) show an understanding that electric current is the rate of flow of charge and solve problems using the equation I = Q/t (b) derive and use the equation I = nAvq for a current-carrying conductor, where n is the number density of charge carriers and v is the drift velocity (c) recall and solve problems using the equation for potential difference in terms if electrical work done per unit charge, WV Q= (d) recall and solve problems using the equations for electrical power P = VI, P = I2R and 2VP R= (e) distinguish between electromotive force (e.m.f.) and potential difference (p.d.) using energy considerations (f) show an understanding of and use the terms period, frequency, peak value and root-mean-square (r.m.s.) value as applied to an alternating current or voltage (g) represent a sinusoidal alternating current or voltage by an equation of the form x = xo sin t (h) deduce that the mean power in a resistive load is half the maximum (peak) power for a sinusoidal alternating current (i) distinguish between r.m.s. and peak values, and recall and use Irms = 2 0I and 2 VV o rms= for the sinusoidal case (j) explain the use of a single diode for the half-wave rectification of an alternating current

2025 Temasek Junior College 2 1 Charge and electric current LO (a)(b) 1.1 Charge The property of matter that causes the electrical force is charge. All bodies are either positively charged, negatively charged, or neutral (zero charge). Bodies with the same charge repel each other and bodies with opposite charge attract. Charge cannot be created or destroyed so therefore charge is conserved. The law of conservation of charge states that in a closed system the amount of charge is constant. Charge is a scalar quantity. The SI unit of charge is the coulomb (C). The charge of electron is e = –1.60 x 10-19 C. Any charge can only exist as an integral multiple of this value. i.e. Q = Ne Examples of charged particles or charged carriers in conductors are • negative electrons in copper wire • positive and negative ions in an electrolyte • positive ions and negative electrons in a discharge tube • negative electrons and positive holes in a semiconductor 1.2 Electric current Electric current I is the rate of flow of charge. i.e. Q t=I The SI unit of current is ampere (A). Hence, in a circuit carrying a constant current I, the charge Q which flows in time t, is given by charge = current x time i.e. Q = It The coulomb is defined as the charge which passes a point in a circuit when a constant current of one ampere flows for one second. If N electrons, each of charge e, pass through a point in a circuit in time t, the total charge is given by Q = Ne and the current Q Ne tt==I . Note: If the current is not constant, but is changing with time, then the instantaneous current is dQ dt=I . And the charge which flows can be found from the area under the current-time graph according to the equation Q dt= I .

2025 Temasek Junior College 3 Example 1: The current in a wire is 200 mA. Calculate (a) the charge which passes a point in the wire in 5 minutes, (b) the number of electrons needed to carry this charge Solution: (a) Q = It = 200 x 10-3 (5 x 60) = 60 C (b) Q = Ne 60 = N (1.60 x 10-19) N = 3.75 x 1020 Example 2: A car battery is used to supply varying current for 5 s. Calculate the total charge delivered from the battery if the variation in current supplied is given by the graph as shown on the right. Solution: Total charge, Q = area under I-t graph Q = 1 2 (3 + 5)  20 = 80 C Example 3: A high p.d is applied between the electrodes of a hydrogen discharge tube so that the gas is ionised. Electrons then move towards the positive electrode and the protons towards the negative electrode. In one second, 5.0 x 1018 electrons and 5.0 x 1018 protons pass a cross- section of the tube. Calculate the current in the discharge tube. Solution: Oppositely-charged particles moving in opposite directions under the influence of an electric field. By convention, current has the same direction as the flow of positive charge. The positive protons moving to the right results in a current Ip to the right. The negative electrons moving to the left results in a current Ie to the right. So the total current is ep 18 18 19 (N N )eQ tt (5.0 x 10 5.0 x 10 )1.6 x 10 1.6 A1 − +== +== I 1 2 3 4 5 6 7 0 10 20 30 I/A t/s low potential high potential Ip Ie

2025 Temasek Junior College 4 1.3 Drift velocity and the derivation of I = nAvq LO (b) A conducting wire is made of metal, such as copper or silver, as shown in Fig. 1.1. Fig. 1.1 In a typical metal, one electron from each atom breaks free to become a free or conduction electron. The atom remains as a positively charged ion. Since there are equal numbers of free electrons (negative) and ions (positive), the metal has no overall charge , i.e. it is neutral. When a battery is connected across the ends of a metal wire in Fig. 1.2, an electric field is set up across the wire. Fig. 1.2 Under the influence of this external electric field, the ‘free’ electrons in the wire experience a force, and accelerate from a region of high potential (i.e. +ve terminal) to a region of low potential (-ve terminal). As the electrons move, it collide with the lattice ions and are scattered repeatedly. The resultant motion is irregular, but it has a slow drift towards the positive terminal with an average velocity called the drift velocity, v. Consider a current I flowing through a section of cylindrical wire of length l and cross-sectional area A as shown in Fig. 1.3. This current is caused by free electrons drifting in the opposite direction to the current. (Note that conventionally current is the flow of positive charges) Fig. 1.3 If there are n electrons per unit volume in the wire with a drift velocity v, then total charge passing through the wire is given by Q = Nq where N is total number of electrons and q is the charge of each electron.

2025 Temasek Junior College 5 Substitute N = n x (Al) where Al is the volume of the wire, Charge Q = n x (Al) q Current I = Q/t I = n x (Al) q/t Sub drift speed v = l/t I = nAvq Note: • Number of charge carriers per unit volume is called number density n. • For metals, the number density, n of charge carriers is approximately 1029 electrons per m3 and drift velocity, v is typically of the order of around 10- 4 m s-1. • When a domestic lighting circuit is switched on, the light comes on almost instantly. This is because the electric field is established almost immediately. So all electrons in the wire and filament bulb move simultaneously under the electric field. Example 4: The number of free electrons per cubic metre in a copper wire is approximately 8 x 10 28. A typical diameter of wire is 1 .0 mm. If the current is 1.0 A, calculate the drift speed of t he electrons along the wire. Solution: ( ) 2 23 28 19 5 -