University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 1
Solutions (Rev B, 2/5/2012)
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. To find the amount of charge, we relate energy to voltage:
Q=
500 J
E
=
= 55.56 C
V
9V
55.56
20
Since each electron has a charge of 1.6 × 10−19 C, we need 1.6×10
−19 = 3.47 × 10
electrons. Because we are charging up the battery, we need to move positive charge
into the positive terminal. Therefore, electrons are flowing in the opposite direction,
from b to a.
2. We can first use dimensional analysis to find the average amount of power consumed
over 30 days.
$0.16
× (30 × 24 hrs) = $115.2/kW
kWh
$30
= 260 W
$115.2/kW
Now we simply apply the power equation to find the average current draw:
I=
260 W
= 2.17 A
120 V
As for your 150 W computer, we already know how much we pay per unit of power
consumed, so we can just scale it by 150 W:
$115.2/kW × 150 W = $17.28
3. We first find the total amount of charge stored on the cell phone battery using
dimensional analysis:
Q = 1650 mAh = (1.65 A) × (3600 s) = (1.65 C/s) × (3600 s) = 5.94 kC
Given this, we can find the energy that the battery is capable of delivering. Since we
have the rate at which energy is dissipated (0.4 W), we can also find how long the
battery will last.
t=
E
Q×V
5.94 kC × 3.7 V
=
=
= 54945 s = 15.3 hrs
P
P
0.4 W
4. The power rating of a light bulb is the power that is produced when connected to the
standard line voltage of 120 V. Thus, bulb A with a rating of 35 W has a resistance of
RA =
V2
(120 V)2
=
= 411.4 Ω
P
35 W
And bulb B with a rating of 85 W has a resistance of
RB =
(120 V)2
V2
=
= 169.4 Ω
P
85 W
Now consider connecting the bulbs in series. They will have the same current, given
by Ohm’s law and the series resistance:
IA = IB =
120 V
= 0.207 A
(411.4 + 169.4) Ω
Applying Ohm’s law to each bulb will give us the voltage:
VA = I × RA = (0.207 A) × (411.4 Ω) = 85 V
VB = I × RB = (0.207 A) × (169.4 Ω) = 35 V
Finally, the power across each bulb can be found using any of the above quantities:
PA = VA × IA = (85 V) × (0.207 A) = 17.6 W
PB = VB × IB = (35 V) × (0.207 A) = 7.25 W
Since bulb A draws more power, it will be brighter than bulb B.
Connecting the bulbs in parallel will leave the same voltage across each one:
VA = VB = 120 V
Ohm’s law can give us the two currents, which are not necessarily equal this time:
IA =
VA
120 V
=
= 0.292 A
RA
411.4 Ω
IB =
VB
120 V
=
= 0.708 A
RB
169.4 Ω
We can find the power as before:
PA = VA × IA = (120 V) × (0.292 A) = 35 W
PB = VB × IB = (120 V) × (0.708 A) = 85 W
Clearly, bulb B will burn brighter in this case. Note that these results should make
sense to you; the respective powers agree with the power rating, as we are connecting
120 V across each bulb.
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5. (a) 0.10 $/kWhr ×
1 hr
3600 s
×
1 kW
1000 J/s
(b) 3.20 $/gal × 264 gal/m3 ×
(c) 0.1 A × 20 hr ×
3600 s
1 hr
= 2.77 × 10−8 $/J
1 m3
720 kg
×
1 kg
4.5×107 J
= 2.61 × 10−8 $/J
$0.75
= 7200 C; 7200 C × 1.5 V = 10800 J; 10800
= 6.94 × 10−5 $/J
J
Gasoline and residential electric power have almost the same cost per joule; AA
batteries are thousands of times more expensive than the other two sources.
6.
Z
∆q =
Z
0
∞
i(t) dt
∞
0.085τ e
(τ ω sin(ωt) − cos(ωt)) −t/τ
−0.085 cos(ωt)e
dt = −
τ 2ω2 + 1
0
τ
= 0 − 0.085 2 2
τ ω +1
= −2.69 × 10−14 C
−t/τ
The battery supplies -2.69 × 10−14 C of charge to the rest of the circuit. The total
energy is thus given by
E = QV = (−2.69 × 10−14 C)(5 V) = 1.35 × 10−13 J
7. When short circuit happens, large amount of current can flow, which causes excessive
heat and a potential fire. The main purpose of a fuse or circuit breaker is to interrupt
this flow of current when a short circuit occurs.
A fuse disintegrates when it heats up to a certain temperature, therefore opening the
circuit in the process. It is a one-time use and would need replacement every time it
burns out.
A circuit breaker, on the other hand, can be reused multiple times. There are many
implementations of a circuit breaker. One implementation is the use of bimetallic
strip. When the bimetallic strip heats up, it will bend and displace a certain
distance. This bending/displacement can then be used to initiate the switch that
trips/opens the circuit.
8. (a) With the two batteries in series with each other and with the load resistor, the
circuit looks like the following:
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The current I can be found by writing a KVL loop. Assume we start at the
bottom left corner and go clockwise, treating voltage rises as positive and
voltage drops as negative:
3 − 10I + 5 − 50I − 100I = 0
Thus the current through the load is I = 0.05 A. The voltage drop can be found
easily using Ohm’s law:
V = IR = (0.05 A)(100 Ω) = 5 V
(b) With the batteries in parallel, the circuit now looks like the following:
This time, it is easier to write a nodal equation for the top node and solve for
the unknown voltage. Note that this voltage is also the same voltage as vL , the
voltage across the load, provided we define the ground on the bottom node as
shown.
vL − 3 vL − 5
vL
+
+
=0
10
50
100
Thus, the voltage drop across the load is 3.077 V, and the current is (again
through Ohm’s law) I = VR = 0.0377 A.
The current through each battery is the same as the currents through their
internal resistances. Thus, the current through the 3 V battery is
I = 3.077−3
= 0.0077 A from the positive to negative terminal. On the other
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hand, the current through the other battery flows in the opposite direction, with
a magnitude of I = 5−3.077
= 0.0385 A.
50
(c) A battery rated at 100 mA ∗ hr holds a charge of
Q = 0.1 × C/s ∗ hr × 3600 s/hr = 360 C
Assuming each battery starts off with that amount of charge when hooked up to
the load, we can use the current through each battery to find the discharge time,
since current is the flow of charge per unit time. Thus, in the series case, both
batteries discharge simultaneously:
t=
Q
360 C
=
= 7200 s
I
0.05 A
4
The parallel case is a bit more complex, since the currents are such that the 5 V
battery is charging the other one. Thus, the former must completely discharge
before the 3 V battery can start charging. Here we assume that while the first
one discharges, the second’s charge content remains constant at 360 C
(essentially a maximum capacity).
t1 =
Q
360 C
=
= 9351 s
I
0.0385 A
Once the first battery discharges completely, it becomes a short circuit, and we
would need to solve for the new current through the 3 V battery. The new nodal
equation is
vL − 3 vL
vL
+
+
=0
10
50 100
This gives us vL = 2.308 V and a current through the battery 0.0692 A. Note
that the battery is now discharging, as the current flows in the opposite
direction. Thus, the time needed to discharge is
t2 =
360 C
Q
=
= 5202 s
I
0.0692 A
Note that this is in addition to the time it took to discharge the first battery.
Note that the 5V battery is now charging. In a real battery, the terminal voltage
would also increase while charging but in our simple model, it would remain at
zero until enough charge is transferred to bring it back to full capacity. In this
example, during time t2 , a total charge of
Q2 = t2 × I2 = t2
vL
= 240C
50
is transferred to the second battery. Since this is below the capacity of the
battery, in our simple model, it does not recharge to 5V.
(d) Connecting two batteries in series is physically viable, and it usually results in
an increased total voltage that is just the sum of the two individual voltages
(minus any losses from internal resistance). However, connecting them in
parallel is often not a good idea unless the cells are well matched. Since the
internal resistance is small, this can potentially generate a huge current
(consider Ohm’s law across the small resistances!), burning out our devices. In
the ideal case, the internal resistance tends to 0, giving us a violation of KVL.
9. KCL at the top left node gives us ia = −ib = −1 A. At the upper right node, we have
id = if − ie = 1.5 A. Finally, we can calculate ic = −ib − id = −2.5 A. Only elements
A and B are in series.
10. ie = ia − ib = −4 A.
if = ie − id = −7 A.
ig = ib + id − ic = 6 A.
Here no elements are in series.
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11. To find voltages, we write KVL equations around the loops in the circuit. For loop
DEB, we have 6x − x − 5 = 0, so x = 1. Thus, vb = 1 V and ve = 6 V. The second
loop, ABC, gives us x + vc − 5 = 0, so vc = 4 V. No elements are in shunt.
12. Here we immediately see that elements C, D, and F are in shunt, so they must all
have the same voltage drop. Thus, vd = vc = 1 V while vf = −1 V. A and E are also
in shunt, giving us ve = va = 3 V. Finally, KVL gives us −va + vb + vc = 0, so
vb = 2 V.
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