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Hints for Exercise 1 on: Current and Resistance Review the concepts of: •  electric current, conven8onal current flow direc8on, •  current density, carrier dri= velocity, carrier number density, •  Ohm’s law, electric resistance, •  power dissipa8on in resistors, Joule hea8ng. As the main source for these concepts, you need to carefully review and re-­‐work your pre-­‐class assignments on “Electric Current” and parts of “Kirchhoff’s Rules”. All ques8ons in Exercise 1 are covered in these two assignments. In addi8on, you may want to consult the corresponding sec8ons in your textbook. Other possible sources to consult: Wikipedia, Google, … Exercise 1 I
I
a
b
A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(a) Find the voltage drop between the two ends of the wire, a and b,
Vab = Va − Vb .
Here, Va and Vb are the electrical potential values at the left end, a, and the right end, b,
respectively. Is Vab > 0 or Vab < 0 ? [Hint: By convention of current flow direction, the
electric current in a wire or other resistor always flows from high to low electric potential.]
(b) Find the amount of heat energy generated in the wire if the current flows for 5 h.
(c) Find the electric field strength, E, inside the wire, and the direction of the E-vector.
Does E point from left to right or from right to left? [Hint: the electric field vector always
points in the direction of decreasing electric potential.]
Exercise 1 (cont’d.) I
I
a b A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(d) The charge carriers in the wire are electrons and the current, I, is due to the average
“drift” motion of these “conduction electrons” inside the wire (see/review FlipIt Physics).
In which direction, on average, do these electrons move: left to right or right to left?
[Hint: An electron has negative charge, q = −e < 0. Consider the direction of the electric
force, F, acting on the electron, due to the electric field, E, found in part (c).]
(e) Find the change in the electric potential energy, ΔU, incurred by a conduction electron,
as it moves from one end of the wire to the other, due to the current flow. Does the
electron gain potential energy (ΔU>0) or lose potential energy (ΔU<0)?
[Hint: Use the potential difference, Vab, found in part (b); or calculate the work done on the
electron by the electric force F , acting on the electron, due to the electric field E ].
Exercise 1 (cont’d.) I
I
a b A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(f) Suppose in some other galaxy, far, far away, the wire was made of anti-matter. The
charge carriers in this “anti-aluminum” wire are then anti-electrons, a.k.a. positrons,
each positron having a positive charge, q = +e > 0. Assuming the same direction and
strength of the electric current, I, the same wire length and diameter, and the same resistivity
as before, answer questions (a)-(e) for the anti-aluminum wire and, respectively, for the
“conduction positrons” flowing through the wire.
Note: You don’t have to re-do any of the numerical calculations. Instead just state for each
part, (a)-(e), whether, or not, the answer for the anti-aluminum wire and its positron charge
carriers is the same as for the aluminum wire with electron charge carriers; and, if not, how
the answer changes. In particular, how do the answers change, or not, for parts (d) and (e)?
Exercise 1 (cont’d.) I
I
a b A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
Returning to our own galaxy and planet earth…
(g) Calculate the total number of electrons, N5h , flowing through the aluminum wire
during 5h. [Hint review the definition of electric current, I, and recall how much
charge, q, each electron carries.]
(h) From N5h found in part (g) and ΔU found in part (e), calculate the combined total
gain or loss of electric potential energy, ΔUtot , from all electrons flowing through the
wire during 5h. Compare ΔUtot to the result from part (b): the total heat energy generated
in 5h by the current flow.
Exercise 1 (cont’d.) I
I
a b A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(i) Calculate the conduction electron number density, ne , in aluminum, that is, the number
of conduction electrons per volume. Use this:
Since aluminum is trivalent (in case you already know any chemistry), each aluminum
atom provides 3 conduction electrons. The molar mass of aluminum is 27 g per mole of
aluminum atoms. The number of aluminum atoms in one mole is given by Avogadro’s
number, NA =6.02×1023 atoms per mole. The mass density of aluminum is 2.70 g/cm3.
(j) Calculate the electron drift speed, that is, the mean velocity of the electrons’ motion
due to the current flow. [Hint: use ne found in part (i), the current I, the wire cross section
and the electron’s charge, q = -e.]
Hints for Exercises 2.1, 2.2 and 3 on: Resistor Circuits Review the rules for parallel and serial combina8on of resistors, summarized on the next page. For more details on these rules, review and re-­‐work your pre-­‐class assignment on “Electric Current” and “Kirchhoff’s Rules” where these rule are discussed. Also read the corresponding sec8on in your textbook. Remember the rules for parallel and serial combina8on of resistors, R1 and R2: Parallel: Voltage Drops: V1 = Va1 – Vb1 , V2 = Va2 – Vb2 , V12 = Va – Vb I1 b1 a1 R1 I12 Currents: I1 = (1/R1) V1 , I2 = (1/R2) V2 I12 a b I2 a2 Equivalent Resistance R12: R12 = V12 / I12 è 1/R12 = 1/R1 + 1/R2 b2 R2 Voltage Drops: V1 = Va – Vb , V2 = Vb – Vc , V12 = Va – Vc Serial: I12 Voltage Drops are Equal: Currents Add Up: V1 = V2 = V12 I1 + I2 = I12 I1 b a R1 I2 Currents: I1 = (1/R1) V1 , I2 = (1/R2) V2 I12 c R2 Voltage Drops Add Up: V1 + V2 = V12 Currents are Equal: I1 = I2 = I12 Equivalent Capacitance R12: R12 = V12 / I12 è R12 = R1 + R2 Addi8onal Hints for Exercise 3 Review and re-­‐work Exercise 3 in the “Capacitance and Capacitor Circuit” lecture notes and especially the step-­‐by-­‐step “Solu8on Hints” posted for that Exercise 3. Then follow these same steps, analogously applied to the resistor circuit shown below. Exercise 3 Part (a) Basic Idea: Simplify the circuit, step by step, by successively replacing parallel or serial resistor combina8ons by their respec8ve equivalent resistances, going from the complicated original circuit, Circuit O, to a final circuit, Circuit 3, which will contain only one single equivalent resistance, connected directly to the bacery: Circuit O à Circuit 1 à Circuit 2 à Circuit 3 Exercise 3 Part (b) Basic Idea: Step by step, going backwards from Circuit 3 à Circuit 2 à Circuit 1 à Circuit O, restore original resistor components from the resp. equivalent resistances. Use the parallel or serial resistor combina8on rules to infer the voltage drops and currents of the original resistor components from the currents and voltage drops across the respec8ve equivalent resistances.