Electrical Quantities and Units

1. Electrical Quantities and Units

2. Electrical Charges (Q) • All matters made of atoms which consist of electrons, protons and neutrons. • Protons and neutrons are in the nucleas and electrons are moving around them. • Each electron carries a negative charge of 1.6 x 10-19 C, each proton carries positive charge of the same value and neutron carries no charge. • Atom having a number of electron not same as proton is called ion.

3. Electrical Charges (Q) • A positive ion is an atom having a number of electron less than number of proton. • A negative ion is an atom having a number of electron more than number of proton. • Coulomb is a unit for electric charge. • 1 coulomb = 6.25 x 1018 electron • Units for electric charge are microcoulomb (C = 10-6C) , nanocoulomb(nC= 10-9, picocoulomb (pC = 10-12 C),

4. Electrical Current (I) • Current is a net flow of electrical charges passing a point and given as i = dq/dt • Unit for current is ampere (A) i.e a rate of 1 C per sec. E.g 230 mA = 230 x 10-3 A = 0.23 A 0.015 A = 0.015 x 103 mA = 15 mA 125 A = 125 x 10-3 mA = 0.125 mA 125 A = 125 x 10-6 A = 0.000125 A

5. Example 1 A q1 Y q2 X The figure shows the flow of electric current. In 2 ms, the values of negative charge q1 and positive charge q2 flowing from X to Y crossing a cross-section at point A are10 C and -4 C respectively. (i)What is the current if both flowing in the same direction (ii)What is the current if both flowing in the different directions (i)The charge net flow is Q = q1 + q2 = 10 C + (-4) C = 6 C. Time t = 2 ms I = Q/t = (6 x 10-6)/(2 x 10-3) = 3 mA (ii)The charge net flow is Q = q1 – q2 = 10 C - (-4) C = 14 C. I = Q/t = (14 x 10-6)/(2 x 10-3) = 7 mA

6. A B q2 = 1 C q1 Potential different (voltage) • Same charge repels . Different charge attracts • Energy is required to bring a positive charge near to negative charge. • Energy is required to bring a positive charge away from another positive charge. • Potential different is a measurement of energy required to bring 1 C charge near to another charge or to bring away same. The unit of potential different is volt (V) and the formula is v = dw/dq • If W = 1 J is required to bring a charge q2 = 1 C from point A to point B, the potential different between A and B is 1 V.

7. Current I = 1 ampere Resistance R = 1 ohm Potential different V = 1 volt Resistance and resistor (R) • The moving of charges in conductor caused collision and friction among them and causing losses of energy. Thus the moving of charges are said to have resistance. • Unit for resistance ohm (). 1 ohm if 1A of current flowing in a conductor to produce 1V between two points • The resistance value depends on material and some other parameter such as temperature

8. Resistance is important element to control the current in the circuit.. • Resistance value is dependent on the following parmeter. • length (l), cross-section (A) and resistivity (), Thus the resistance an be writen as R = l/A or R = 1/(A)  = 1/ = conductivity • Resistance also depends on temperature and can be fomulated as R1 = temperature coef. Of resistance at t1 R2 = temperature coef. Of resistance at t2 0 = pekali suhu rintangan pada suhu 0 C • Most insulator such as rubber , the resistance decreases with the increases of temperature.

9. Typical values of resistivity

10. Typical temperature coefficients of resistance referred to 0 deg C

11. Example 2 For cuprum:  = 0.0173 -m. If the length of the cuprum wire is 10 m and its cross-section is 0.5 mm2 . What is its resistance? R = l/A = (0.0173 x 10-6 x 10)/(0.5 x 10-6) = 0.346  = 346 m

12. Example 3 A cable consists of two conductors which , for the purposes of a test , are connected together at one end of the cable. The combined loop resistance measured from the other end is found to be 100 W when the cable is 700m long. Calculate the resistance of 8 km of similar cable.

13. Example 4 A conductor of 0.5 mm diameter wire has a resistance of 300 W. Find the resistance of the same length of wire if its diameter were double.

14. Example 5 • A coil of copper wire has a resistance of 200W when its mean temperature is 0oC. Calculate the resistance of the coil when its mean temperature is 80oC

15. Example 6 • When a potential difference of 10V is supplied to a coil of copper wire of mean temperature 20oC, a current of 1 A flows in the coil. After some time the current falls to 0.95 A yet the supply voltage remains unaltered. Determine the mean temperature of the coil given that the temperature coefficient of resistance of copper is 4.28 x 10-3 / oC at 0oC At 20oC At q2oC

16. Ohm’s law Potential different (V) across the resistor is proportional to current (I) Thus we can write as V  I (A graph V-I is linear and then V/I is a constant) This constant is called a resistance (R): Hence V = RI E.g. A simple dc circuit consist of voltage source V=10V and resistor producing a current of 4 mA. What is the value of resistance R? From Ohm’s law: R = V/I = 10/(4 x 10-3) = 2.5 k

17. I Rv 1 M (rheostat) R Rp 10 k 1.2 k (fixed resistor) (potentiometer) Simple Circuit & symbols

18. + V - ri + V - Voltage source DC voltage source Ideal source Practical source

19. ~ v(t) AC voltage source • Generate a variable potential difference with time • The variation of potential different follows a sine waveform, so the voltage varied in amplitude and phase which can represented by V (t) = A sin (wt + q) where A is amplitude, w = 2p f , f is frequency , t time and q is phase displacement

20. + + ix vx V V - - (a) Current dependent source (b) Voltage dependent source Dependence voltage source • The voltage source depends on other parameter such as input current or input voltage of a device. The symbols are as follows

21. I + i(t) Current source Direct current (DC) Alternating current (AC)

22. ix vx I I (a) Dependence on current (b) Dependence on voltage Dependence Current source • The current source which depends on other parameter such as input current or input voltage of a device. The symbols are as follows

23. Energy For physical quantities W= F(Newton) x d (m) W= P t F=ma For Electrical energy W= I2 R t W= VI t Where F =force, d=distance, t = time , a= acceleration, P=power, m= mass , u=velocity, V=voltage, I=current and R=resistance

24. Example 7 • A current of 3 A flows through a 10 W resistor. • Find: • The power developed by the resistor • (b) the energy dissipated in 5 min. (a) (b)

25. Example 8 A heater takes a current of 8 A from a 230 V source for 12 h. Calculate the energy consumed in kilowatt hours kWh is the unit used in determine amount of energy used in electricity

26. Power (P) Physical quantities In case of rotating electrical machine Power efficiency T= torque , Nr = rotation speed (revolution per min), nr=revolution per sec, Po= output power, Pin =input power

27. Power (P) • Power is the rate of consuming the energy (The rate of work done) • For (ac) current the power in rms : P = VI From Ohm’s law V = IR  P = I2R or from Ohm‘s law; I = V/R  P = V2/R • Unit for power is in watt (W) • A power of 1 W is a rate of energy consuming for 1 J per sec. Thus 1 W = 1 J/s

28. Example 9 An electric motor has a torque of 48 Nm at rotation speed of 1800 revolution per minute (r.p.m). The efficiency of the motor is 88%. If the power factor is 0.85, calculate the current drawn by the motor when it is connected to the main supply of 415 V. and

29. Example 10 • A 230 V lamp is rated to pass a current of 0.26 A. Calculate its power output. If a second similar lamp is connected in parallel to the lamp, calculate the supply current required to give the same power output in each lamp. Power for parallel lamps

30. R = 2 M I = 10 A Example 11 • Calculate the dissipated power from a resistor of 2MW when a current of 10 mA flowing in it. • P = I2R = (10 x 10-6)2 x 2 x 106 • = 200W

31. Current (A) masa (s) • For current and voltage which are not constant, the power must be calculated as follow : • Divide into time interval and the instantaneous power are • Rms =(root mean square )