33-228
Electronics
Spring 2016
Version of January 19, 2016
Carnegie Mellon University
Department of Physics
Homework 1
SOLUTIONS
Problems 1.4, 1.5, 1.10, 1.15, 1.17, 1.20, 1.D42, 1.D48 and 1.D51
1.4 (a) The resistance of a conductor is given as
R =
l
ρ.
A
For our copper wire, we have that ρ = 17.2×10−9 Ωm, l = 50 m and A = 2.08×10−5 m2 .
Thus, the resistance of our wire is 0.041 Ω. (b) If 15 A of current flows through the
wire, then the voltage drop along the wire is ∆V = IR, or 0.62 V . (c) For 16-gauge
wire, we have that the area is A = 5.24 × 10−6 m2 , so the resistance of the wire will be
R = 0.16 Ω and the voltage drop is 2.46 V .
1.5 We want to have the same resistance for both pieces of wire. Given that
R =
l
ρ,
A
we need that
l
l
ρcu =
ρal .
Acu
Aal
We can solve this for the area of the aluminum wire to find that
Aal = Acu
ρal
.
ρcu
We can now write this in terms of diameters to yield that
s
dal = dcu
ρal
.
ρcu
This gives us that the minimum diameter of the aluminum wire must be 2.08 mm.
If we could get it, this would require 11-gauge wire. However, it is likely that only
10-gauge is readily available.
1.10 For two inductors, L1 and L2 , in series, the total voltage across them is
V
= V1 + V2 .
1
Since both inductors must have the same current through them, the rate of change of
the current, dI/dt must also be the same in each of them. Thus, we have that
dI
dt
dI
V2 = L2
dt
Hence, we must have that Leq = L1 + L2 .
V1 = L1
1.15 This is a simple application of Ohm’s law, V = IR. We are told that V = 10 V and
R = 10 kΩ, so we have that
V
,
R
or putting in numbers, we get that I = 1 mA.
I =
1.17 The Thévenin equivalent circuit is shown in Figure 1.
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Vth
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‚
Rth
A
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ò
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B
Figure 1: The Thévenin equivalent circuit for problem .
1.20 ⊗ Given the resistances of the two meters, we can replace the two meters with their
known resistances and then compute voltages an currents. In the case where the
voltmeter is across both the resistor and the ammeter, we measure the voltage V0 .
However, to get the total current, we have these resistance 1.01R in parallel with
100R, which is R to three significant figures. This means that the total current is
V0 /R, but it is divided with a small portion going through the voltmeter, V0 /100R, or
1%. This means that the ammeter will measure a current of 0.99V0 /R.
In the second case, the voltmeter is across the resistor, so the resistor and voltmeter in
parallel will yield an equivalent resistance of 0.99R, which when added to the resistance
of the ammeter gives a total circuit resistance of R. The total current (which now goes
through the ammeter) is V0 /R. However, there is a small voltage drop across the
ammeter of 0.01V0 which means that the voltmeter must measure 0.99V0 .
D1.42 We can either place resistors in series or in parallel. The basis of this problem is to
understand what the tolerance of these two combinations are, and then see of we can
come up with a circuit that can reduce the tolerance to 10%.
Let us start with the series combination, where we will take two resistor, R1 and R2 .
The equivalent resistance is
Req = R1 + R2 ,
2
and the error is this relative to the first resistor is found by taking a derivative,
dReq
= 1
dR1
dReq = dR1 ,
so, combining errors, we have
dReq =
h
dReq
=
Req
"
(dR1 )2 + (dR2 )2

i1
2
(dR1 )2
(dR2 )2
+
(R1 + R2 )2 (R1 + R2 )2
dReq
dR1
= 
Req
R1
!2 R1
R1 + R2
2
#1
2
dR2
+
R2
!2 R2
R1 + R2
2
1
2

Now let us consider the case where R1 = R2 and dR1 = dR2 , then we can rewrite the
expression as

1
dReq
= 
Req
4

1
dReq
= 
Req
2
dR1
R1
dR1
R1
!2
1
+
4
dR1
R1
!2  12

!2  12

1 dR1
dReq
= √
Req
2 R1
√
So placing resistors in series can reduce the relative error by 1/ n where n is the
number of resistors in the chain. Thus, if we took four 150 Ω resistors in series, each
with a 20% tolerance, the equivalent resistor would be 600 Ω with a tolerance of 10%.
We also need to consider two resistors in parallel. Again, if we start with R1 and R2 ,
then we have
Req =
R1 R2
.
R1 + R2
We can take the derivative with respect to R1 to get that
dReq
R2
=
dR1
R1 + R2
dReq
R1 R2
=
dR1
R1 + R2
dReq
R1 R2
=
dR1
R1 + R2
3
R1 R2
(R1 + R2 )2
1
1
−
R1 R1 + R2
1
1
−
R1 R1 + R2
−
dReq
1
R2
= Req
dR1
R1 R1 + R2
dReq
R2
dR1
=
.
Req
R1 + R2 R1
We would arrive at a similar expression for R2 , in that
dReq
R1
dR2
=
,
Req
R1 + R2 R2
so that we have

dReq
R2
= 
Req
R1 + R2
2
dR1
R1
!2
R1
+
R1 + R2
2
dR2
R2
!2  12

.
Now let us assume that R1 = R2 and that dR1 = dR2 , we then have

dReq
1
= 
Req
4
dR1
R1
!2
1
+
4
dR1
R1
!2  12

.
dReq
1 dR1
= √
.
Req
2 R1
√
Thus, the error of n resistors in parallel will be reduced by 1/ n, as with the resistors
in series. Thus, if we had 4 2.4 kΩ resistors, we could place them in parallel and achieve
our desired result.
D1.48 The Thévenin voltage and resistance tell us how external components will behave when
connected to the circuit, but they tell us nothing about how the circuit itself behaves.
For example, if it is a voltage divider, then current will flow through the resistors in
the divider even when no load is connected. What is true is that no power will be
dissipated external to the circuit if no load is connected.
D1.51 The rational for the two resistors in parallel is presumably to limit the power through
either of them. Taking the nominal values of all the components, we find the total
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€ …€ÿ‚ ÿ‚…€ ò
‚
„ƒ
„ƒ
3.3kΩ
3.3kΩ
„ÿƒ€‚ ò
„ƒ€ ‚€
„ÿƒ
‚
„ƒ
25 V
440Ω
Figure 2: The circuit from problem D1.36.
4
current through the circuit is
V
Rtot
25 V
I =
2090 Ω
I ≈ 12 mA
I =
The power dissipated in the 440Ω resistor is
P440 = (12 mA)2 (440Ω)
P440 = 63 mW
which is well under the 125 mW limit of the resistor. For the two in parallel, the
nominal current through each will be 21 of our 12 mA, so the
P3300 = (6 mA)2 (3300Ω)
P3300 = 118 mW .
This is close, but under the 18 W limit of the resistors. However, let us assume that
the resistors are not the exact value, but rather some value within the 20% tolerances.
Doing the calculations, we find the following power dissipations for up to 20% changes
in resistor values as listed in Table 1. For a number of combinations in which values
are more than 10% from nominal values, we start to exceed the power ratings of the
components. Thus, while nominal values are protected, unless we carefully choose our
resistors, we can exceed power ratings of our components and risk device failure.
PA
RA
RB
3300Ω 3300Ω 118mW
3200Ω 3400Ω 121mW
3100Ω 3500Ω 125mW
3000Ω 3600Ω 129mW
2640Ω 3960Ω 145mW
PB
118mW
115mW
111mW
108mW
97mW
Table 1: Power dissipation with varied resistor values in the circuit for problem D1.51.
5