PHYSICS 1142
Semester 2, 2010
Electromagnetism: Assignment #2 - Electricity.
SOLUTIONS
1
An axon membrane of a nerve cell is approximately a cylinder of length 12 mm and
diameter 150 µm. The membrane has a capacitance of 1.0µF cm-2. There is a resting
potential difference of -90 mV between the inside and outside of the cell.
[Hint: area of cylinder radius r, length l is A = 2 π r l ]
a)
How much electrical potential energy is stored in the membrane?
Total area:
Total capacitance:
Total energy:
A = 2 π r l = 2 × π × 75 × 10-6 × 0.012 = 5.65 × 10-6 m2
C = 5 × 10-6 × 5 × 10-6 × = 5.65 × 10-8 C
1
U =
C V 2 = 2.3 × 10-10 J
2
-10
ENERGY =____ 2.3 × 10
J ___________
[3 marks]
b) How much charge is there on each surface? Give the answer in units of the elctron
charge (e = 1.6 × 10-19 C).
Q = C V = 5.65 × l0-8 × 0.090 = 5.1 × 10-9 C
Q = n e ⇒ n = 3.2 × 1010
CHARGE =____
2
5.1 × 10-9 C ___________
[2 marks]
Consider the circuit below.
a) Calculate the potential difference across the 25 Ω resistor.
1
25 Ω in parallel with 100 Ω is equal to:
= 20 Ω
1
1
+
25 100
20
8
12 =
voltage is therefore:
= 2.7 V
20 + 70
3
POTENTIAL DIFFERENCE =______
2.7 V _____
[3 marks]
b) Calculate the current through the 25 Ω resistor.
I
=
V
R
=
2.7
= 0.11 A
25
CURRENT =______
0.11 A ______
[2 marks]
3
A length of wire is cut into three equal lengths, which are then wrapped together
side by side to make a thicker wire with one third of the initial length. How does the
resistance of this new combination compare to the resistance of the original wire?
!L
Ro =
originally
A
L
!
3 = Ro
R =
finally
3A
9
NEW R
OLD R =_____
4
At $0.16 per kWh what does it cost to leave a 60 W porch light on all day for a year?
rate
cost = 0.16
×
kW
60 × 10-3
×
hours in year
365 × 24 = 84.096
COST =______
5.
1/9 ______
[3 marks]
$ 84.10 _____
[2 marks]
Calculate the current that flows through the 100 Ω resistor.
Let i1 and i2 be the currents through each battery:
By symmetry i1 = i2 , so current through R is 2i1.
Sum of voltages around left-hand loop:
Vo ! i1 r ! 2i1 R = 0
⇒
i1 =
Vo
r + 2R
=
9.0
= 44.6 mA
2 + 2 ! 100
So through R current is 89 mA
CURRENT =____
89 mA ____
[3 marks]
6
How many 600 W electric kettles, operated at 240Vrms, can be used without blowing an
8 Arms fuse?
600 n ≤ 240 × 8
⇒
n ≤ 3.2
therefore adding a 4th kettle (or anything else!) will blow the fuse
NUMBER =_______ 3 _______
[2 marks]
7
Suppose a current is given by the equation I = 2.30 sin 90t, where I is in amperes and t
in seconds. (a) What is the frequency? (b) What is the rms value of thc current? (c) If
this is the current through a 100.0 Ω resistor, what is the equation that describes the
voltage as a function of time?
I = Io sin 2π f t = 2.30 sin 90t
90
f = 2π
Irms =
1
2
(a) FREQUENCY =_______
Io
=
2.30
2
14.3 Hz _______
[2 marks]
(b) CURRENT =______
1.63 A ______
[2 marks]
V = R I , Vo = R Io = 100.0 x 2.30 = 230 V
8
(c) VOLTAGE =
V = 230 sin 90 t
[2 marks]
Suppose you want to run some apparatus that is 75 m from an electric outlet. Each of
the wires connecting your apparatus to the 100 V source has a resistance per unit length
of 0.0055 Ω/m. If your apparatus draws 2.5 A, what will be the voltage drop across the
connecting wires and what voltage will be applied to your apparatus?
Across each wire V = R I =
for both wires
ρ
(A) l I = 0.0055 x 75 x 2.5 = 1.03
2.1 V _______
[2 marks]
APPARATUS VOLTAGE =________ 98 V _________
[2 marks]
9
VOLTAGE DROP =________
A heart pacemaker is designed to operate at 60 beats/min using a 4.0 µF capacitor.
What value of resistance should be used if the pacemaker is to fire when the voltage
reaches 63 percent of maximum?
- t/τ
for capacitor charging
V = Vmax ( 1 - e
)
where τ = RC
1
for t = τ
V = Vmax ( 1 - e ) ≈ 0.63 Vmax
this must occur once every second (60 beats/min) hence τ = RC ≈ 1 s
τ
1
R = C = 4.0 x 10-6 = 2.5 × 105
RESISTANCE =_______ 250 kΩ _______
[4 marks]
RNJ 8-Sept-10