Download Electricity Tutorial Sheet 1 2010

Foundation Electric Science
Tutorial Sheet 1
2010/11
Tutorial 1
for
Chapter 1 Electrical Energy and Cell
Question 1-1:
If 2000 C electrons travel in a wire in 1000 s, what is the current flowing in the wire in the
units A and mA respectively ?
Answer:
Q=2000 C, t=1000 s, so
I
Q 2000

 2 A.
t 1000
Since 1A=1000 mA,
I  2  1000  2000 mA.
Question 1-2:
As shown in Figure for Question 1-2, a closed circuit consists of a cell and two resistors of
equal resistance in series. E=15 V and VB=0 V. Find the potentials at the point A and C
respectively ; and find the potential differences across AB, BC and AC.
-
+
I
E=15 V
VAB
VBC
C
VC
A
B
VB
VA
Figure for Question 1-2
Answer:
VAC= E=15 V.
|VAB |= | VBC | = 0.5 x 15 = 7.5 V.
Since VB= 0 V,
VA= 7.5 V.
VC= - 7.5 V.
VAB = 7.5 V.
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Foundation Electric Science
Tutorial Sheet 1
2010/11
VBC = - 7.5 V.
Question 1-3:
If a cell can generate 1000 C electrons by consuming chemical energy of 1500 J, what is the
voltage across the circuit?
Answer:
Electrical energy is equal to the amount of chemical energy, 2000 J.
V
W 1500

 1.5 V.
Q 1000
Question 1-4:
240 V is powered to the DC motor that consumes electric energy of 2 kJ per second when it
is in use. Find the electric current to travel in the motor.
Answer:
We know P  VI , hence
I
P 2000

 8.3 A.
V
240
Question 1-5:
A new cell of E 1.8 V is connected to an electrical device. The measured voltage across the
device is 1.5 V. What is the lost voltage VL.
Answer:
VL  E  1.5  1.8  1.5  0.3 V.
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Foundation Electric Science
Tutorial Sheet 1
2010/11
Question 1-6 :
The open circuit voltage across Cell 1is 15 V and that across Cell 2 is 15 V. They are
connected in series in a closed circuit of resistors. If we set the potential of zero at the
connection, then what is the voltage across Cell 1 and what is the voltage across Cell 2 in the
circuit?
Cell 2
Vc
-
Cell 1
+
Vb
-
Vbc
+
Va
Vab
Figure for Question 1-6
Answer:
The voltage across Cell 1 is +15 V, and the voltage across Cell 2 is -15 V.
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Foundation Electric Science
Tutorial Sheet 1
2010/11
Question 1-7:
An electron has a charge of 1.6 × 10-19 C, and a mass of 9.11 × 10-31 kg. It is held stationary
in an electric field generated in the capacitor. The electron is released and allowed to
accelerate through a potential difference of 2 kV. Calculate the final speed of the electron.
Answer:
An electron is negative and so it will be forced to move from the negative terminal where the
potential is Va to the positive terminal where the potential is Vb in the electric field.
Electrical energy Ee will be converted to mechanical energy Em in acceleration of the
electron, in which the speed of the electron will be changed from zero to the maximum speed.
Electron Q
-
+
v
-
+
-
+
Electrical field
Vab
Va
Vb
Figure for Question 1-7
Step 1：Find the electrical energy
The electrical energy is the product of the voltage and the amount of electrical charge.
Ee  W  VabQ  2000  1.6  10 19  3200  10 19 J .
Step 2：Find the mechanical energy
Em 
1 2
mv , where m is the mass and v is the final speed of the electron.
2
Step 3：Find the final speed of the electron
v
2 Em

m
2  3200  10 19
 7  1014  2.7  10 7 ms 1 .
31
9.11 10
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Foundation Electric Science
Tutorial Sheet 1
5
2010/11