Foundation Engineering Science Electric Circuits: Tutorial Sheet 1
2011/12
Chapter 1 Electrical Energy and Cell
Tutorial 1: Solutions
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. V = 15V and VB = 0V. Find the potentials at the point A and C
respectively; and find the potential differences across AB, BC and AC.
-
+
I
V =15V
VBC
C
VC
VAB
A
B
VB
VA
Figure for Question 1.2
Answer:
VAC = V = 15 V.
Since VB = 0V,
VAB = VBC = 7.5V.
VA = +7.5V.
VC = –7.5V.
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Foundation Engineering Science Electric Circuits: Tutorial Sheet 1
2011/12
Question 1.3:
If a cell can generate 1000 C electrons by consuming chemical energy of 2000 J, what is the
voltage across the circuit?
Answer:
Electrical energy is equal to the amount of chemical energy, 2000 J.
V
W 2000

 2.0V
Q 1000
Question 1.4:
240V is powered to the DC motor that consumes electric energy at a rate of 2 kJ per second
when it is in use. Find the electric current that flows through the motor.
Answer:
We know P  VI , hence
I
P 2000

 8.33 A
V
240
Question 1.5:
A new cell with internal voltage, V = 1.8 V is connected to an electrical load resistance. The
measured voltage across the load VR = 1.5 V. What is the lost voltage VL.
Answer:
VL  V  1.5  1.8  1.5  0.3 V.
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Foundation Engineering Science Electric Circuits: Tutorial Sheet 1
2011/12
Question 1.6:
The open circuit voltage across Cell 1 is 10V and across Cell 2 is also 10V. They are
connected in series in a closed circuit of resistors. If we set the potential of zero at the
connection (i.e. VB), then what is the voltage at VA relative to VB and what is the voltage at VC
relative to VB?
Cell 1
VC
-
Cell 2
+
VB
-
VBC
+
VA
VAB
Figure for Question 1.6
Answer:
The voltage at VA relative to VB = +10V and the voltage at VC relative to VB = –10V.
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Foundation Engineering Science Electric Circuits: Tutorial Sheet 1
2011/12
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 negatively charged, so it will be repelled from the negative plate where the
potential is Vb and attracted to the positive plate where the potential is Va 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
Vb
Va
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  3.2 10 16 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
2  3200  10 19

 7  1014  2.7  10 7 ms 1 .
m
9.11 10 31
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