chapter 1 exercises - test bank direct · chapter 1 exercises 1.1 the current isi = (1 v) =(1 ) = 1...

Chapter 1 Exercises

1.1The current is I = (1 V)/(1 Ω) = 1 A. 1 A is 1 C/s, and the electron charge has magnitude

1.6× 10−19 C, so the number of electrons per second is (1/1.6× 10−19) /s = 5.2× 1018/s.

1.2a) From (1.2.8) we have

R =mL

e2nAτ.

Solving for τ gives

τ =mL

q2nAR.

To convert units we note that 1 Ω = 1 V/A and 1 C-V = 1 J = 1 kg-m2/s2.

(9.1× 10−31 kg)(.002 m)

(1.6× 10−19 C)2(1021/cm3)(100 cm/m)3(.001 m)2(100 V/A)' 7× 10−19 s.

where we used the standard mass and charge of an electron. This is quite short, because thedensity given in the problem of 1021 cm−3 is unrealistic—that would be a typical density fora highly conducting metal. The electron density of the resistor should have been given as1015 cm−3, which would give τ = 7× 10−13 s.

b)-c) Using the classical equipartition theorem 12mv2 = 3

2kBT , the average velocity is

v =√

3kBT/m =√

3(1.38× 10−23 J/K)(300 K)/(9.1× 10−31 kg) = 1.2× 105 m/s.

For the electron density given in the problem, the mean free path is then

l = vτ = 8.2× 10−14 m.

This is much smaller than the typical spacing of atoms in a crystal on the order of a fewangstroms (1 A = 10−10 m). A more realistic electron density and scattering time, e.g. 1015

cm−3 as intended for this problem, would give the mean free path much longer than theatomic spacing.

1.3The mobility is

µ =qτ

m,

so that

τ =mµ

q=

(9.1× 10−31 kg)(10−2 m2/V − s)

(1.6× 10−19 C)= 6× 10−14 s.

By the same classical calculation of v as in Exercise 1.2, the mean free path is 7.2× 10−9 m.

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Solution Manual for Electronics A Physical Approach 1st Edition by Snoke

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1.4Poisson’s equation in spherical coordinates is

2∂V

∂r+∂2V

∂r2= 0.

which has the solution

V (r) = −V0

r

ab

b− a+ V0

b

b− a.

The electric field is

E(r) = −V0

r2

ab

b− a,

and

J = σE = −V0σ

r2

ab

b− a.

The surface area of the sphere at radius b is 4πb2, so the current is

I = JA =V0σ

b2

ab

b− a(4πb2) = V0σ

4πab

b− a

and therefore

R =V0

I=

1

4πσ

b− aab

.

1.5 The ratio of output to input gives

R2

R1 +R2

= 0.4.

The current (assuming no load resistance) is I = (10 V)/(R1 + R2) ≤ 1 mA, and thereforeR1+R2 ≥ 104 Ω. Solving for the equality with R1+R2 = R2/0.4 gives R1 = 6000, R2 = 4000.

1.6 The ratio of output to input with no load gives

R2

R1 +R2

= 0.2.

From (1.3.5), when there is a load resistance, we have

VoVi

=R2

R1 +R2 +R1R2/RL

= (0.95)(0.2) = 0.19.

Solving the two equations for R1 and R2 with RL = 100 gives R1 = 26.3, R2 = 6.6.

1.7Loops:

2






V1 = I1R1 + I2R2

V1 = I1R1 + ImRm + I3R3

V1 = I1R1 + ImRm + I4R4 + V2 .

Nodes:

I1 = I2 + ImIm = I3 + I4.

R1 R2 0 0 0R1 0 Rm R3 0R1 0 Rm 0 R4

−1 1 1 0 00 0 −1 1 1

I1

I2

ImI3

I4

=

V1

V1

V1 − V2

00

Matrix equation solver gives

I1 = − (−R2R3V1 −R2R4V1 −R3R4V1 −R3RmV1 −R4RmV1 +R2R3V2)

(R1R2R3 +R1R2R4 +R1R3R4 +R2R3R4 +R1R3Rm +R2R3Rm +R1R4Rm +R2R4Rm)

I2 = − (−R3R4V1 −R3RmV1 −R4RmV1 −R1R3V2)


Im = − (−R2R3V1 −R2R4V1 +R1R3V2 +R2R3V2)


I3 = − (−R2R4V1 −R1R2V2 −R1RmV2 −R2RmV2)


I4 = − (−R2R3V1 +R1R2V2 +R1R3V2 +R2R3V2 +R1RmV2 +R2RmV2)


The voltage drops are just given by ∆Vn = InRn for each resistor.

1.8Loops:

V = I1R1 + I3R3 + I6R6 + I7R7

V = I1R1 + I2R2 + I4R4 + I6R6 + I7R7

V = I1R1 + I2R2 + I5R5 + I6R6 + I7R7

Nodes:

I1 = I2 + I3

I2 = I4 + I5

I3 + I4 + I5 = I6

I6 = I7.

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1.9The matrix and solution are the same as for Exercise 1.7. The relevant equation is

Im = − −R2R3V1 −R2R4V1 +R1R3V2 +R2R3V2

R1R2R3 +R1R2R4 +R1R3R4 +R2R3R4 + (R1R3 +R2R3 +R1R4 +R2R4)Rm

.

The current for the Thevenin equivalent circuit is

Im =1

Rm +RTh

VTh,

which gives

RTh =R1R2R3 +R1R2R4 +R1R3R4 +R2R3R4

R1R3 +R2R3 +R1R4 +R2R4

and

VTh =−R2R3V1 −R2R4V1 +R1R3V2 +R2R3V2

R1R3 +R2R3 +R1R4 +R2R4

.

1.10Current for voltage source:

I =VTh

RTh +RL

.

Current through load for current source:

I = ITh −V

Ri

= ITh −IRL

Ri

,

I =ITh

1 +RL/Ri

,

If ITh = VTh/Ri, this becomes

I =VTh/Ri

1 +RL/Ri

=VTh

Ri +RL

which is the same as the voltage source current if Ri = RTh.

1.11Pick the current to give 1 V reading when R is at the maximum value, e.g. for R = 100,

pick I = 1 V/100 Ω = .01 A = 10 mA. Can read with three digits of accuracy in each case.

1.12For upper loop:

ImRm = IL(R2 +RL +R3)

4






Node equation:

I = IL + Im

Solve:∆Vo = ILRL = (IRmRL)/(R2 +R3 +RL +Rm).

When RL →∞, ∆Vo → IRm.RL should be about a million times larger than the other resistances.

1.13A solution is shown in Figure 1.

Figure 1: Geometry of a 4-lead measurement, for Exercise 1.13.

1.14

1

R1/Rv + 1=

Rm/Rv

R2/Rv +Rm/Rv

(R2/Rv +Rm/Rv) = (Rm/Rv)(R1/Rv + 1)

5






1.15 Assume current flows from top to bottom, and from left to right across Ri:Loops:

V = I1R1 + IvRv

V = I2R2 + ImRm

V = I1R1 + IiRi + ImRm

Nodes:

I1 = Ii + IvI2 + Ii = ImUsing an equation solver, we obtain

Ii = − (R1Rm −R2Rv)V

R1R2Ri +R1R2Rm +R1RiRm +R1R2Rv +R2RiRv +R1RmRv +R2RmRv +RiRmRv

.

As expected this vanishes when (1.7.2) is fulfilled.

1.16Current of 1 mA for 2 hours corresponds to

Q = (.001 A)(2× 3600 s) = 7.2 C.

This is N = (7.2 C)/(1.6 × 10−19 C) = 4.5 × 1019 electrons, or half this many Zn atomsremoved.

Mass of Zn is 65(1.7× 10−27 kg)(4.5× 1019) = 5× 10−6 kg.

1.17The total current is

I =V

R1 +R2

.

The power dissipations in the resistors are

P1 = I2R1 =R1

(R1 +R2)2V 2

P2 = I2R2 =R2

(R1 +R2)2V 2.

The higher resistance burns out first. Setting the burn out power to 0.25 W gives

V = (R1 +R2)√P2/R2 = (300 Ω)

√0.25V · A/(200 Ω) = 10.6 V.

6






1.18

P =V 2

R.

R =V 2

P=

(10 V)2

.25 W= 400 Ω.

A lower resistance with the same power rating will burn out.

1.19

I = V/(R1 +R2).

P1 = I2R1 =R1V

2

(R1 +R2)2≤ 0.25 W.

Solving this for R2 gives

R2 ≥√R1V 2

P1

−R1 =

√√√√(100 Ω)(10 V)2

(0.25 W)− 100 Ω = 100 Ω.

1.20Loops:

V = I1R1 + I3R3 + I1(R6 +R7)V = I1R1 + I2R2 + V + I1(R6 +R7)

Node:

I1 = I2 + I3.

Solution:

I1 =R2V

R1R2 +R1R3 +R2R3 +R2R6 +R3R6 +R2R7 +R3R7

I2 = − (R1 +R6 +R7)V

R1R2 +R1R3 +R2R3 +R2R6 +R3R6 +R2R7 +R3R7

I3 =(R1 +R2 +R6 +R7)V

R1R2 +R1R3 +R2R3 +R2R6 +R3R6 +R2R7 +R3R7

.

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1.21Loops:

Vi = I1R1 + I3R3 + VoVi = I1R1 + I2R2 + V + VoVo = I6R6.

Nodes:

I1 = I2 + I3

I3 + I2 = I6.

Vo =R6(R2Vi +R3Vi −R3V )

R1R2 +R1R3 +R2R3 +R2R6 +R3R6

1.22Loop:

V = I2R2 + I3R3

Node:

I + I2 = I3

I2 = −IR3 − VR2 +R3

I3 = I − IR3

R2 +R3

+V

R2 +R3

.

1.23 Solving the circuit gives

I1 =R2

R1R2 +R1R3 +R2R3

Vi = (0.0014/Ω)Vi

I2 =R1

R1R2 +R1R3 +R2R3

Vi = (0.014/Ω)Vi

I3 =R1 +R2

R1R2 +R1R3 +R2R3

Vi = (0.015/Ω)Vi.

P1 = I21R1 = (0.0014/Ω)2(50 Ω)V 2

i = (0.0001 Ω)V 2i

8






P2 = I22R2 = (0.014/Ω)2(5 Ω)V 2

i = (0.001 Ω)V 2i

P3 = I23R3 = (0.015/Ω)2(60 Ω)V 2

i = (0.014 Ω)V 2i

Assuming that all the resistors have the same power rating, R3 burns out first. At thispoint no current flows through the circuit.

1.24 This exercise has a typo: for the assignment of R1 and R2 in Figure 1.9, the choice ofresistor values given in the exercise will give Vo/Vi = 2/3. We will do the solution assumingthat 2/3 is the correct value for Vo/Vi.

From (1.3.5), when there is a load resistance, we have

VoVi

=R2

R1 +R2 +R1R2/RL

=20

10 + 20 + (10)(20)/(50)= 0.588Vi

instead of the designed (2/3)Vi.The voltage divider could use resistors with the same ratio of resistances but lower resis-

tance, e.g. 10 kΩ and 20 kΩ, which would give

VoVi

=20

10 + 20 + (10)(20)/(50, 000)

which deviates from 2/3 only in the 4th decimal place.

1.25A current source can be made simply from a voltage supply in series with a resistance

Ri. The total current for the internal resistor and the load resistor is

I =V

Ri +RL

.

We want I when RL = 100 Ω to be 99% of I when RL = 0:

V/(Ri +RL)

V/Ri

= 0.99.

Solving for Ri gives

Ri =0.99

0.01RL = 99RL = 9900 Ω.

A resistance greater than this will give less than 1% drop in current when RL is increasedfrom 0 to 100 Ω.

1.26 A straightforward current source would be a 10-V voltage supply with a 1-kΩ resistor,which gives 100 mA flow through the resistor. Resistance in parallel with the 1-kΩ resistorwould not give significant droop if the resistance is large compared to 1 kΩ; resistance inseries would not give significant droop if the resistance is small compared to 1 kΩ.

9






For N identical resistors in parallel,

I = NV

R,

and the minimum resistance is given by

R =V 2

Pmax

,

which implies

I = NV

V 2/Pmax

= NPmax

V.

Thus by increasing N the current can always be increased. Thus for an ideal voltage supplythere is no upper bound to the amount of current that can be drawn.

1.27The circuit is shown in Fig. 2.

Figure 2: Circuit for Exercise 1.27.

Loops:

Vi = I1R1 + I2R2

Vi = I1R1 + I1′R1′ + VoVo = I1′R2′

Node:I1 = I2 + I1′

Solution:VoVi

=R2R2′

R1R1′ +R1R2 +R1′R2 +R1R2′ +R2R2′.

10






This is not the same as the product of two voltage divider relations:

R2

R1 +R2

R2′

R1′ +R2′=

R2R2′

R1R1′ +R1′R2 +R1R2′ +R2R2′.

1.28 One simple solution is to wire each bulk in parallel with a resistor with comparableresistance, as shown in Figure 3:

Figure 3: Christmas light circuit for Exercise 1.28.

A drawback is that the circuit uses more power, as power is lost in the parallel resistors,but if a bulb blows out, the other bulbs will still light.

1.29The circuit shown in Figure 4 is effectively the same as Figure 1.10 in the book. Without

RDyer, the current through RLight is

I =V

R1 +RLight

.

With the dryer connected, the current through the lights is

IL =V

R1 +RLight +R1(RLight/RDryer)

Thus there is “droop” in the voltage across RLight.

11







1.30A solution is shown in Figure 5.


1.31a) The output voltage is zero, since there is no current flow through R4 or R5 (or R3),

and therefore no voltage drop across those resistors, so that both output connections sharethe same voltage, which is equal to that at the node between R1 and R2.

b) This would not change if an resistor is added between the contacts.c) If the low side is connected to ground, then the output voltage is effectively measured

across R5. Solving for this voltage we have

Loops:

Vi = I1R1 + I3R3 + I3R5

Vi = I1R1 + I2R2

12






Vo = I3R5

Node:I1 = I2 + I3

Solution:

Vo =R2R5

R1R2 +R1R3 +R2R3 +R1R5 +R2R5

Vi

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chapter 1 exercises - test bank direct · chapter 1 exercises 1.1 the current isi = (1 v) =(1 ) = 1...

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