Hints for Exercise 1 on: Current nd Resistnce Review the concepts of: electric current, conventionl current flow direction, current density, crrier drift velocity, crrier numer density, Ohm s lw, electric resistnce, power dissiption in resistors, Joule heting. As the min source for these concepts, you need to crefully review nd re-work your pre-clss ssignments on Electric Current nd prts of Kirchhoff s Rules. All questions in Exercise 1 re covered in these two ssignments. n ddition, you my wnt to consult the corresponding sections in your textook. Other possile sources to consult: Wikipedi, Google,
Exercise 1 A current = 2.5 A is flowing from left to right through stright luminum wire with circulr cross section of dimeter D=(0.3 / π 1/2 ) mm nd length L=5.0 m, s shown ove (not drwn to scle!). Aluminum hs resistivity of 2.7 10-8 Ω m (t room temperture). () Find the voltge drop etween the two ends of the wire, nd, V = V V. Here, V nd V re the electricl potentil vlues t the left end,, nd the right end,, respectively. s V > 0 or V < 0? [Hint: By convention of current flow direction, the electric current in wire or other resistor lwys flows from high to low electric potentil.] () Find the mount of het energy generted in the wire if the current flows for 5 h. (c) Find the electric field strength, E, inside the wire, nd the direction of the E-vector. Does E point from left to right or from right to left? [Hint: the electric field vector lwys points in the direction of decresing electric potentil.]
Exercise 1 (cont d.) A current = 2.5 A is flowing from left to right through stright luminum wire with circulr cross section of dimeter D=(0.3 / π 1/2 ) mm nd length L=5.0 m, s shown ove (not drwn to scle!). Aluminum hs resistivity of 2.7 10-8 Ω m (t room temperture). (d) The chrge crriers in the wire re electrons nd the current,, is due to the verge drift motion of these conduction electrons inside the wire (see/review Flipt Physics). n which direction, on verge, do these electrons move: left to right or right to left? [Hint: An electron hs negtive chrge, q = e < 0. Consider the direction of the electric force, F, cting on the electron, due to the electric field, E, found in prt (c).] (e) Find the chnge in the electric potentil energy, ΔU, incurred y conduction electron, s it moves from one end of the wire to the other, due to the current flow. Does the electron gin potentil energy (ΔU>0) or lose potentil energy (ΔU<0)? [Hint: Use the potentil difference, V, found in prt (); or clculte the work done on the electron y the electric force F, cting on the electron, due to the electric field E ].
Exercise 1 (cont d.) A current = 2.5 A is flowing from left to right through stright luminum wire with circulr cross section of dimeter D=(0.3 / π 1/2 ) mm nd length L=5.0 m, s shown ove (not drwn to scle!). Aluminum hs resistivity of 2.7 10-8 Ω m (t room temperture). (f) Suppose in some other glxy, fr, fr wy, the wire ws mde of nti-mtter. The chrge crriers in this nti-luminum wire re then nti-electrons,.k.. positrons, ech positron hving positive chrge, q = +e > 0. Assuming the sme direction nd strength of the electric current,, the sme wire length nd dimeter, nd the sme resistivity s efore, nswer questions ()-(e) for the nti-luminum wire nd, respectively, for the conduction positrons flowing through the wire. Note: You don t hve to re-do ny of the numericl clcultions. nsted just stte for ech prt, ()-(e), whether, or not, the nswer for the nti-luminum wire nd its positron chrge crriers is the sme s for the luminum wire with electron chrge crriers; nd, if not, how the nswer chnges. n prticulr, how do the nswers chnge, or not, for prts (d) nd (e)?
Exercise 1 (cont d.) A current = 2.5 A is flowing from left to right through stright luminum wire with circulr cross section of dimeter D=(0.3 / π 1/2 ) mm nd length L=5.0 m, s shown ove (not drwn to scle!). Aluminum hs resistivity of 2.7 10-8 Ω m (t room temperture). Returning to our own glxy nd plnet erth (g) Clculte the totl numer of electrons, N 5h, flowing through the luminum wire during 5h. [Hint review the definition of electric current,, nd recll how much chrge, q, ech electron crries.] (h) From N 5h found in prt (g) nd ΔU found in prt (e), clculte the comined totl gin or loss of electric potentil energy, ΔU tot, from ll electrons flowing through the wire during 5h. Compre ΔU tot to the result from prt (): the totl het energy generted in 5h y the current flow.
Exercise 1 (cont d.) A current = 2.5 A is flowing from left to right through stright luminum wire with circulr cross section of dimeter D=(0.3 / π 1/2 ) mm nd length L=5.0 m, s shown ove (not drwn to scle!). Aluminum hs resistivity of 2.7 10-8 Ω m (t room temperture). (i) Clculte the conduction electron numer density, n e, in luminum, tht is, the numer of conduction electrons per volume. Use this: Since luminum is trivlent (in cse you lredy know ny chemistry), ech luminum tom provides 3 conduction electrons. The molr mss of luminum is 27 g per mole of luminum toms. The numer of luminum toms in one mole is given y Avogdro s numer, N A =6.02 10 23 toms per mole. The mss density of luminum is 2.70 g/cm 3. (j) Clculte the electron drift speed, tht is, the men velocity of the electrons motion due to the current flow. [Hint: use n e found in prt (i), the current, the wire cross section nd the electron s chrge, q = -e.]
Hints for Exercises 2.1, 2.2 nd 3 on: Resistor Circuits Review the rules for prllel nd seril comintion of resistors, summrized on the next pge. For more detils on these rules, review nd re-work your pre-clss ssignment on Electric Current nd Kirchhoff s Rules where these rule re discussed. Also red the corresponding section in your textook.
Rememer the rules for prllel nd seril comintion of resistors, R 1 nd R 2 : Prllel: 1 1 1 R 1 12 12 2 2 2 R 2 Voltge Drops: V 1 = V 1 V 1, V 2 = V 2 V 2, V 12 = V V Currents: 1 = (1/R 1 ) V 1, 2 = (1/R 2 ) V 2 Voltge Drops re Equl: V 1 = V 2 = V 12 Currents Add Up: 1 + 2 = 12 Equivlent Resistnce R 12 : R 12 = V 12 / 12 è 1/R 12 = 1/R 1 + 1/R 2 Seril: 1 12 12 R 1 2 R 2 c Voltge Drops: V 1 = V V, V 2 = V V c, V 12 = V V c Currents: 1 = (1/R 1 ) V 1, 2 = (1/R 2 ) V 2 Voltge Drops Add Up: V 1 + V 2 = V 12 Currents re Equl: 1 = 2 = 12 Equivlent Resistnce R 12 : R 12 = V 12 / 12 è R 12 = R 1 + R 2
V Bt = 5V
V Bt = 5V
Additionl Hints for Exercise 3 Review nd re-work Exercise 3 in the Cpcitnce nd Cpcitor Circuit lecture notes nd especilly the step-y-step Solution Hints posted for tht Exercise 3. Then follow these sme steps, nlogously pplied to the resistor circuit shown elow. Exercise 3 Prt () Bsic de: Simplify the circuit, step y step, y successively replcing prllel or seril resistor comintions y their respective equivlent resistnces, going from the complicted originl circuit, Circuit O, to finl circuit, Circuit 3, which will contin only one single equivlent resistnce, connected directly to the ttery: Circuit O à Circuit 1 à Circuit 2 à Circuit 3 Exercise 3 Prt () Bsic de: Step y step, going ckwrds from Circuit 3 à Circuit 2 à Circuit 1 à Circuit O, restore originl resistor components from the resp. equivlent resistnces. Use the prllel or seril resistor comintion rules to infer the voltge drops nd currents of the originl resistor components from the currents nd voltge drops cross the respective equivlent resistnces.
V Bt = 5V