Module M2-1 Electrical Engineering

Transcription

1 Module M-1 Electrical Engineering TUTOIL 4 Topics Conductors (ต วนำไฟฟ า) Dielectrics or insulators (ฉนวนไฟฟ า) Capacitance CONDUCTOS ND DIELECTICS CPCITNCE SEPTEME 1, 016 fter this tutorial, you will be able to State the relationship of the current density and the current for the case that the current density is uniform and perpendicular to the conductor s cross-section Calculate resistance of a conductor with a uniform cross-section pply Ohm s law for cases of one resistor State the relationship of ~D and E ~ within a dielectric 3 (continued) Compute the equivalent resistance of resistors connected in series or parallel Calculate capacitance of a parallel-plate capacitor Calculate the total energy stored in a parallel-plate capacitor Compute the equivalent capacitance of capacitors connected in series or parallel 4

2 Example 1: Properties of conductors Conductors 5 = d =mm ~E 6 -mm-diameter copper wire with conductivity of S/m is subject to a electric field of 0 mv/m ssume that the electric field is uniform and perpendicular to the cross-section of the wire Question 1.1: Find the current density ~E 7 ~ J I Question 1.: Find the current flowing in the wire = d =mm ~ J 8 Solution: The current density is given by The direction of The magnitude of is the same as the direction of ~ J ~ E ~ J is ~ J = ~ E J = E = /m Solution: The electric field is uniform and perpendicular to the cross-section, so the current is I = JS = S/m V/m From question 1.1, J = /m cross-section area of a circle is d 10 3 S = = = m

3 Question 1.3: The wire is 50-m long; determine the wire s resistance 9 Question 1.4: Determine the voltage V (potential difference) across the wire s length 10 L = 50 m I V ~E V I Solution: esistance of a conductor of uniform crosssection is 50 = L =0.7 S cross-section area (calculated in Question 1.4) Solution: Ohm s law gives us V = I 3.64 (from question 1.) 0.7 (from question 1.3) Or use a formula of the potential difference due to uniform ~E : V = V V = EL =1V V =1V Question 1.5: Find the power dissipated in the wire Solution: Power dissipated in a conductor is P = I 11 =3.64 W Example : resistors connected in series and parallel 1 I =3.64 (from question 1.) 0.7 (from question 1.3) Each resistor is = 4 Find the equivalence resistance between points and

4 Solution: epeatedly find the equivalence resistance 1 = 1 eq + 1 = 3 series 13 eq = + = 3 series eq = + 3 = 5 3 parallel (next page) 5 3 (continued) Solution 14 1 = 1 eq = 8 5 parallel The equivalence resistance between points and is 5 8 = 15 = Example : Dielectrics Dielectrics 15 ~E out ~ Ein 16 elative permittivity of Teflon is r = free space Teflon piece of Teflon is subject to an external uniform electric field ~E out = b i V/m The electric field inside the Teflon is measured to be ~E in =0.5 b i V/m

5 Question.1: Find the electric flux density in free space (outside of Teflon) Solution: The electric flux density in free space is 17 Question.: Find the electric flux density within the Teflon Solution: The electric flux density within Teflon is 18 ~D out = 0 ~ Eout = b i C/m ~D in = 0 r ~ Ein = b i C/m permittivity of free space: F/m b i V/m F/m 0.5 b i V/m Example 3: Parallel-plate capacitor Capacitance 19 0 S =5cm d =cm r =6 (relative permittivity of mica) parallel-plate capacitor has a mica dielectric, r = 6, a plate area of 5 cm, and a separation of cm

6 Question 3.1 Calculate the capacitance Solution: The capacitance of a parallel-plate capacitor is = 0 r =( ) 6 1 = F/m S =5cm = 5 (10 m) = m Question 3. If a battery of V is connected to the plates, determine the total energy stored in the capacitor Solution: The total energy stored in the capacitor is C = S d = =1.33 pf W E = 1 CV = J 10 m V = (from question 3.1) Example 4: nother parallel-plate capacitor 3 1 cm 3 cm Solution: The given structure is a parallel connection of two capacitors 1 cm 3 cm 4 5 cm cm 5 cm cm C 1 C r =6 r =4 r =6 r =4 capacitor consists of two parallel plates separated by a distance of cm Two dielectrics are used to fill the space one with r =6 between the rectangles of size 1 cm by 5 cm the other with r =4 between the rectangles of 3 cm by 5 cm Determine the capacitance of this capacitor capacitance C 1 =1.33 pf from question 3. capacitance C = ( ) 4 ( ) 10 =.66 pf The equivalent capacitance of two parallel capacitors is C eq = C 1 + C from the lecture same approach as question 3. s = 3.99 pf

7 Summary Excises related to conductors Current density Current esistance of a conductor with uniform cross-section Ohm s law Power dissipated in a conductor Exercises related to dielectrics elative permittivity Electric flux density inside the material 5 (continued) Exercises related to capacitance Capacitance of a parallel-plate capacitor Energy stored in a capacitor 6 In-class exercise = d =1mm Homework 3, problem 1 8 L = m 7 I =3 copper: conductivity = S/m cylindrical copper wire has a diameter of 1 mm, a length of m, and carries a current of 3 1. Find the total resistance of the wire.. What is the voltage V across the wire s length? V