30: Resistance from Microscopic Ohm's Law

INTRO:
Your task is to calculate the resistance of a simple cylindrical resistor with wires connected to the ends, such as the carbon composition resistors that are used on electronic circuit boards. Imagine that the resistor is made by squirting material whose conductivity is σ into a cylindrical mold with length L and cross-sectional area A as shown in (Figure 1) . Assume that this material satisfies Ohm's law. (It should if the resistor is operated within its power dissipation limits.)

Figure 1

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PART A:
What is the resistance R of this resistor?
Express the resistance in terms of variables L, A and σ given in the introduction. Do not use V, I, E or J in your answer.

SOLUTION:
There are two equations that are necessary to complete this problem. 

Eq. (30.18) states that resistivity ρ = 1/σ

and

Eq. (30.22) states that resistance R = ρL/A

plugging (30.18) into (30.22) gives... 

R = L/σA

NOTE

Real resistors vary tremendously in overall size. The larger the size, the more power the resistor can dissipate without heating to the point that it is dangerous to nearby components or that the material of which it is constructed begins to change its conductivity (i.e., so that the resistance would no longer be constant). The amount of resistance is determined by the conductivity of the material of the resistor, which can vary over more than 20 orders of magnitude. Commercially available resistors vary from 0.1 ohm or less to more than 10⁷ ohm.

30: Problem 30.66

INTRO:
Household wiring often uses 2.0-mm-diameter copper wires. The wires can get rather long as they snake through the walls from the fuse box to the farthest corners of your house.
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PART A:
What is the potential difference across a 16-m-long, 2.0-mm-diameter copper wire carrying an 7.8 A current?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
Using eq. (30. 21)... I = A⋅ΔV/ρL

→ ΔV = IρL/A
​

using table (30.2) → ρ_copper = 1.7×10^-8 Ωm
& A = π/4⋅d²

→ ΔV = (7.8 A)(1.7×10^-8 Ωm)(16 m)/(π/4⋅d²)
⇒ ΔV = 0.675 V = 0.68 V​

30: Problem 30.55

INTRO:
The two wires in the figure (Figure 1) are made of the same material.

Figure 1

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PART A:
What is the current in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Since both segments are connected in series, current is the same 
I₁ = I₂ = 2 A

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PART B:
What is the electron drift speed in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Using eq. (30.13)... J = I/A = n_e⋅e⋅v_d
rearranging... I = A⋅n_e⋅e⋅v_d
I₁ = I₂ 

→ A₁⋅n_e1⋅e⋅v_d1 = A₂⋅n_e2⋅e⋅v_d2​

The electron charge cancels out, and also the intro says that the wires are the same material, so n_e1= n_e2 ... 

→ A₁⋅v_d1 = A₂⋅v_d2
→v_d2 = A₁⋅v_d1 / A₂

= (π/4⋅d₁²)(v_d1) / (π/4⋅d₂²)
= (d₁²)(v_d1) / (d₂²)
= (1×10^-3 m)²(2×10^-4 m/s) / (2×10^-3 m)²
⇒ v_d2 = 5×10^-5 m/s

= 50 μm/s​

30: Problem 30.51

INTRO:
A hollow metal sphere has inner radius a, outer radius b, and conductivity σ. The current I is radially outward from the inner surface to the outer surface.
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PART A:
Find an expression for the electric field strength inside the metal as a function of the radius r from the center.
Express your answer in terms of the variables I, σ, r, and appropriate constants.

SOLUTION:
If you were to consider a thin radial section of the cylinder (so that the cross section would be almost uniform throughout), what would be its resistance? 

The radial area, not cross-sectional this time, is A = 2πrL
the L = 2r ∴ A = 4πr²

J = I/A = σE 

→ E = I/(σA) = I/(σ4πr²)

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PART B:
Evaluate the electric field strength at the inner surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(1×10^-2 m)²]
⇒ E = 3.45×10^-4 V/m

They want E = 3.4×10^-4 V/m though... for some reason.  It'll automatically correct it though.

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PART C:
Evaluate the electric field strength at the outer surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(3×10^-2 m)²]
⇒ E = 3.83×10^-4 V/m

They want E = 3.8×10^-4 V/m ... At least that rounding makes sense :) 

30: Problem 30.45

INTRO:
The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 4.30 mm in diameter and 1.2 m long. The starter motor runs for 0.630 s until the car engine starts.
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PART A:
How much charge passes through the starter motor?
Express your answer with the appropriate units.

SOLUTION:

Givens/ Conversions:

I = 180 A
L = 1.2 m
d = 4.30 mm = 4.3×10^-3 m
t = 0.630 s

Using eq. (30.10): Q = I⋅Δt...

Q = (180 A)(0.630 s)
⇒ Q = 113.4 C (A⋅s = C)

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PART B:
How far does an electron travel along the wire while the starter motor is on?
Express your answer with the appropriate units.

SOLUTION:
Using eq. (30.13) ... J = I/A = n_e⋅e⋅v_d
rearranging... v_d = I/(A⋅n_e⋅e)

this equation has all knowns and solves for the speed. first step for determining distance traveled over time, right? right.
​

using table (30.1) ... n_ecopper = 8.5×10²⁸ m^-3

We know that the distance traveled = velocity⋅time
∴ distance = I⋅Δt/(A⋅n_e⋅e)

= (180 A)(0.63 s)/[(π/4⋅(4.3×10^-3 m)²)(8.5×10²⁸ m^-3)(1.602×10^-19C)]
= 5.735×10^-4 m
⇒ 0.574 mm

30: Problem 30.32

INTRO:
The terminals of a 0.70 V watch battery are connected by a 70.0-m-long gold wire with a diameter of 0.200 mm .

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PART A:
What is the current in the wire?
Express your answer using three significant figures.

SOLUTION:

Givens/ Conversions: ​

V = 0.70 V
L = 70.0 m
d = 0.200 mm = 2×10^-4 m
also, Ω = V/A
​

So.. we have a voltage, a length, a diameter, and the material. 
We know that the cross-sectional area : A = π/4⋅d²
From table (30.2), Gold has a resistivity of: 

ρ = 2.4×10^-8 Ωm​

By using eq. (30.22), we can determine the resistance.

R = ρL/A = ρL/(π/4⋅d²)​

Finally, we know V=IR, by know, that means that I = V/R

or... I = V/[ρL/(π/4⋅d²)] = πVd²/(4ρL)
→ I = π(0.70 V)(2×10^-4 m)²/[4(2.4×10^-8 Ωm)(70.0 m)]
⇒ I = 0.0131 A​

They want it in mA, though. 1 mA = 10^-3 A

⇒ I = 13.1 mA​

30: Problem 30.24

INTRO:
The two segments of the wire in the figure (Figure 1) have equal diameters but different conductivities σ₁ and σ₂. Current I passes through this wire.

Figure 1

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PART A:
If the conductivities have the ratio σ₂/σ₁=5, what is the ratio E₂/E₁ of the electric field strengths in the two segments of the wire?

- 1/25
- 5
- 25
- 1/5

SOLUTION:
Since the current I passes through the entire wire, and the diameters are equal... the current density for both sides is... 
J = I/A

This means that J₁ = J₂

using eq. (30.17) ... J = σE ...

J₁ = σ₁E₁
J₂ = σ₂E₂​

since J₁ = J₂... σ₁E₁ = σ₂E₂

by rearranging...
σ₂/σ₁ = E₁/E₂ = 5 (given)​

E₂/E₁ = [E₁/E₂]^-1

⇒ E₂/E₁ = 1/5​