Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

An aluminum wire with a cross-sectional area of 4 x 10^{- 6} m ² carries a current of 5 A. Find the drift speed of the electrons in the wire, assuming each aluminum atom contributes on electron. The density of aluminum is 2.7 g/cm ³.

Solution:
For this we will use

I = nqv_dA.

To find the number of charges per unit volume, n , we find from the periodic table that the atomic mass of aluminum is 26.98 g/mol. Since 1 mol of a substance contains Avagadro's number N_A = 6.02 x 10²³ of particles, we find

n = 6.02 x 10²³$${\frac{{\:\rm electrons}}{{\:\rm mole}}}$$ x $${\frac{1\ {\:\rm mole}}{26.98\ {\:\rm g}}}$$ x $${\frac{2.7\ {\:\rm g}}{1\ {\:\rm cm}{}^3}}$$ x $$\left(\frac{100\ {\:\rm cm}}{1\ {\:\rm m}}\right)$$³ = 6.02 x 10²⁸$${\frac{{\:\rm electrons}}{{\:\rm m}{}^3}}$$

We then find

v_d = $${\frac{I}{nqA}}$$ = $${\textstyle\frac{5}{6.02\times 10^{28}\cdot 1.6\times 10^{-19}\cdot 4\times 10^{-6}}}$$ = 1.3 x 10^{- 4} m / s .

Thus, the average drift speed is about 0.13 mm/s.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose one wants to make a 0.5 $\Omega$ resistor out of 1 g of Copper. If the resistor is a uniform cylinder, what is the diameter and length required?

Solution:
We will relate the resistance R to the resistivity by

R = $$\rho$$$${\frac{L}{A}}$$,

where for Copper $\rho$ = 1.7 x 10^{- 8} $\Omega$ $\cdot$ m . Since Copper has a density of 8.95 g/cm ³, we know that the volume V of Copper present is

V = $${\frac{1\ {\:\rm g}}{8.95\ {\:\rm g}/\ {\:\rm cm}{}^3}}$$ x $$\left(\frac{1\ {\:\rm m}}{100\ {\:\rm cm}}\right)$$³ = 1.12 x 10^{- 7} m ³.

Since V = LA , we then find

R = $$\rho$$$${\frac{L}{A}}$$ = $$\rho$$$${\frac{L^2}{V}}$$ $$\Rightarrow$$ L = $$\sqrt{\frac{RV}{\rho}}$$ = = $$\sqrt{\frac{ 0.5\cdot 1.12\times 10^{-7}}{ 1.7\times 10^{-8}}}$$ = 1.8 m .

We can then also find the diameter d required as

A = $${\frac{V}{L}}$$ = $${\textstyle\frac{1}{4}}$$$$\pi$$d ² $$\Rightarrow$$ d = $$\sqrt{\frac{4V}{\pi L}}$$ = $$\sqrt{\frac{4\cdot 1.12\times 10^{-7}}{\pi \cdot 1.8}}$$ = 2.8 x 10^{- 4} m .

Thus, the wire must have a length of 1.8 m and a diameter of 2.8 x 10^{- 4} m.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A small motor draws a current of 2 A from a 120 V line. Assuming 100% efficiency, what is the cost of operating the motor for 10 hours if the power company charges $0.050/kWh?

Solution:
The motor has a power output of

P = VI = 2 x 120 = 240 W .

Operating it for 10 hours will consume an amount of energy

E = Pt = 240 W x 10 h x $${\frac{1 \ {\:\rm kW}}{1000 \ {\:\rm W}}}$$ = 2.4 kWh .

It will thus cost 2.4 x $0.05=$.12 to operate the motor.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Find the current through R₄ in the circuit in Fig. 17.9 if V = 30 V, R₁ = 12 $\Omega$ , R₂ = 18 $\Omega$ , R₃ = 9 $\Omega$ , and R₄ = 6 $\Omega$ .

  

Figure 17.9: Resistors in series and parallel

Solution:
We first reduce the four resistors to one equivalent resistance. The resistors R₁ , R₂ , and R₃ are in parallel; they have an equivalent resistance

$${\textstyle\frac{1}{R_p}}$$ = $${\textstyle\frac{1}{R_1}}$$ + $${\textstyle\frac{1}{R_2}}$$ + $${\textstyle\frac{1}{R_3}}$$ = $${\textstyle\frac{1}{18}}$$ + $${\textstyle\frac{1}{9}}$$ + $${\textstyle\frac{1}{6}}$$ $$\Rightarrow$$ R_p = 3 $$\Omega$$.

This equivalent resistance is then in series with R₁ ; the equivalent resistance of these two resistors in series is

R = R₁ + R_p = 12 + 3 = 15 $$\Omega$$.

Thus, the equivalent resistance of the four resistors is 15 $\Omega$ . The current flowing through this equivalent circuit is thus

I = $${\frac{V}{R}}$$ = $${\textstyle\frac{30}{15}}$$ = 2 A ,

from which we deduce the potential difference across the equivalent resistance R_p = 3 $\Omega$ as

V_p = IR_p = 2 x 3 = 6 V .

This potential difference is the same across R₂ , R₃ , and R₄ , and in particular we have through R₄

I₄R₄ = 6 $$\Rightarrow$$ I₄ = $${\textstyle\frac{6}{6}}$$ = 1 A .

Thus, the current through R₄ is 1 A.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Find the power lost in the 50 $\Omega$ resistor in the circuit in Fig. 17.10.

  

Figure 17.10: Circuit illustrating Kirchhoff's laws

Solution:
We begin by labelling the unknown currents as indicated. Applying Kirchhoff's junction rule, we find

I₁ = I₂ + I₃.

We next use Kirchhoff's loop rule; for the top loop, going clockwise, we find

30I₃ + 50I₃ - 40I₂ = 0 $$\Rightarrow$$ I₃ = $${\textstyle\frac{1}{2}}$$I₂,

while for the bottom loop, also going clockwise,

40I₂ - 20 + 10I₁ = 0 $$\Rightarrow$$ I₁ = 2 - 4I₂.

Substituting these last two relations into the first yields

2 - 4I₂ = I₂ + $${\textstyle\frac{1}{2}}$$I₂ $$\Rightarrow$$ I₂ = $${\textstyle\frac{4}{11}}$$ A ,

from which follows

$${\textstyle\frac{1}{2}} {\textstyle\frac{2}{11}}$$

$${\textstyle\frac{6}{11}}$$

The fact that all three currents came out positive indicates the directions we assumed for them were correct.

As a check on our algebra, we can verify that by going around the outer loop

30I₃ + 50I₃ - 20 + 10I₁ = 0,

as it should. We then know the current through the 50 $\Omega$ resistor to be I₃ = ${\frac{2}{11}}$ A, from which we find the power lost as

P = I ²R = $$\left(\frac{2}{11}\right)$$² x 50 = 1.65 W .

Thus, the 50 $\Omega$ resistor dissipates 1.65 W of power.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Find the currents through all three resistors in the circuit in Fig. 17.11.

  

Figure 17.11: Another circuit illustrating Kirchhoff's laws

Solution:
We begin by labeling the unknown currents as indicated. Applying Kirchhoff's junction rule, we find

I₁ + I₃ = I₂.

We next use Kirchhoff's loop rule; for the top loop, going clockwise, we find

20 - 100I₂ - 20I₃ = 0 $$\Rightarrow$$ I₃ = 1 - 5I₂,

while for the bottom loop, also going clockwise,

100I₂ - 60 + 10I₁ = 0 $$\Rightarrow$$ I₁ = 6 - 10I₂.

Substituting these last two relations into the first yields

(1 - 5I₂) + (6 - 10I₂) = I₂ $$\Rightarrow$$ I₂ = $${\textstyle\frac{7}{16}}$$ A ,

from which follows

$${\textstyle\frac{19}{16}}$$

$${\textstyle\frac{13}{8}}$$

In this case the value obtained for I₃ is negative, which indicates that our initial guess for the direction of I₃ was wrong. We also see by this that the 20 V battery is in fact being charged.

As a check on our algebra, we can verify that by going around the outer loop

- 20I₃ + 20 - 60 + 10I₁ = 0,

as it should. Note that for this check we must explicitly use the negative value found for I₃ .