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MIT 8 02 - Electric Currents - Worked Examples

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8.02 Fall, 2002VI. Electric Currents - Worked ExamplesMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall, 2002 VI. Electric Currents - Worked Examples Example 1: Resistivity of a cable The first telegraphic messages crossed the Atlantic Ocean in 1858, by a cable 3000 km long laid between Newfoundland and Ireland. The conductor in this cable consisted of seven copper wires, each of diameter 0.73 mm, bundled together and surrounded by an insulating sheath. (a) Calculate the resistance of the conductor. Use 3160 cm−×Ω⋅ for the resistivity of the copper. (b) A return path for the current was provided by the ocean itself. Given that the resistivity of seawater is about 25 ohm-cm, show that the resistance of the ocean return would have been much smaller than that of the cable. Solution: (a) The resistance R of a conductor is related to the resistivity ρ by /RlAρ=, where l are A are the length of the conductor and the cross-sectional area, respectively. Since the cable consists of N=7 copper wires, the total cross sectional area is 222(0.073cm)744dANr Nπππ== = (1.1) The resistance then becomes ()()()862310cm 310 cm307197 0.073cm / 4lRAρπ−−××Ω⋅== = Ω132 (1.2) (b) Let the current be spread out over an area of A , and the distance across the Atlantic Ocean be approximately l . With the resistivity of seawater being 310 km 1km=10 cm≈×82000 km 2 10 cm≈=×25 cmsρ=Ω⋅ , the resistance of the ocean is ()()8413 225 cm 2 10 cm'510 cm10RR−Ω⋅ ×==×Ω<< (1.3) Thus, the resistance of the ocean return is much smaller than that of the cable. 1Example 2: Charge at junction Show that the total amount of charge at the junction of the two materials in the figure below is 02111(Iε)σσ− , where I is the current flowing through the junction, and1σ and 2σ are the conductivities for the two materials. Solution: In a steady state of current flow, the normal component of the current density J must be the same on both sides of the junction. Since GJEσ=, we have 11 2 2EEσσ= (2.1) or 122Eσσ=1E (2.2) Let the charge on the interface be Qin, we have, from the Gauss’s law: ()210inSQdEEAε⋅=− =∫EAGGv or 210inQEEAε−= (2.3) Substituting using Equation (2.2) then yields 2E 101 01122111inQAE AEσεεσ1σσσ =−= −   (2.4) Since the current is ()11IJA E Aσ== , the above expression becomes 02111inQIεσσ=− (2.5) 2Example 3: Drift velocity The resistivity of seawater is about 25 ohm-cm. The charge carries are chiefly Na and ions, and of each there are about 31 per c . If we fill a plastic tube 2 meters long with seawater and connect a 12-volt battery to the electrodes at each end, what is the resulting average drift velocity of the ions, in cm/sec? +-Cl200×3m Solution: The current in a conductor of cross sectional area A is related to the drift speed vd of the charge carriers by dIenAv= (3.1) where n is the number of charges per unit volume. We can then rewrite the Ohm’s law as ()dlV IR neAv nev lAρdρ== = (3.2) which yields dVvne lρ= (3.3) Substituting the values, we have ()()()()20 -3 195512V6 10 cm 1.6 10 C 25 cm 200cmVcm cm2.5 10 2.5 10Csdv−−−=××Ω⋅⋅=× =×⋅Ω (3.4) In converting the dimensions we have used 1VV1amperesCCC−===Ω⋅ Ω 3Example 4: Circuit Consider the circuit shown below, for a given resistance0R, what must be the value of1R so that the input resistance between the terminals is equal to 0R? Solution: The equivalent resistance, R’, due to the three resistors on the right is ()01101101211 1'RRRRRR RRR+=+ =++ (4.1) or ()10 10'2RRRRRR+=+ (4.2) Thus, the equivalent resistance due to the four resistors becomes ()210 111101 013222eqRR R0RRRRRRRRR++=+ =++ (4.3) If 0eqRR= , then ()200 1 1 10 0 1232 322RRR R RR R R+=+ ⇒= (4.4) or 013RR = (4.5) 4Example 5: Show that, if a battery of fixed emf ε and internal resistance iR is connected to a variable external resistance R, the maximum power is delivered to the external resistor when iRR= Solution: Using Kirchhoff’s rule, ()iIRRε=+ (5.1) which implies iIRRε=+ (5.2) The power dissipated is equal to ()222iPIR RRRε==+ (5.3) To find the value of R which gives out the maximum power, we differentiate P with respect to R and set it to be 0: ()() ()2222120iii iRRdP RdRRR RR RRεε−=−=++ +3= (5.4) which implies iRR= (5.5)


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MIT 8 02 - Electric Currents - Worked Examples

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