Electrical Engineering Science 1-2002 2003 Exam-Electrical Engineering, Exams of Electrical Engineering

Professor Miller, Manchester Metropolitan University, Electrical Engineering, Electrical Engineering Science 1, 2002 2003 Exam, time constant,Thevenin's Theorem , reactance , impedance, power dissipation,full-wave rectifier, RMS value

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2010/2011

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THE MANCHESTER METROPOLITAN UNIVERSITY
FACULTY OF SCIENCE AND ENGINEERING
DEPARTMENT OF ENGINEERING AND TECHNOLOGY
SESSION 2002/2003
Examination for the
HND ELECTRONIC ENGINEERING
YEAR ONE
UNIT 64EE1201 : ELECTRICAL ENGINEERING SCIENCE I
Thursday 8 May 2003
2.00 pm to 4.00 pm
Instructions to Candidates
Attempt ALL questions.
Assessment will be based on the marks obtained from your best TWO attempts.
Breakdown of marks for each question is shown in square parentheses.
All questions carry equal marks.
Students are permitted to use their own calculators subject to Faculty conditions.
Alpha-numeric memories must be cleared prior to the start of the examination.
S275 08/03/04
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THE MANCHESTER METROPOLITAN UNIVERSITY

FACULTY OF SCIENCE AND ENGINEERING

DEPARTMENT OF ENGINEERING AND TECHNOLOGY

SESSION 2002/

Examination for the

HND ELECTRONIC ENGINEERING

YEAR ONE

UNIT 64EE1201 : ELECTRICAL ENGINEERING SCIENCE I

Thursday 8 May 2003

2.00 pm to 4.00 pm

Instructions to Candidates

Attempt ALL questions.

Assessment will be based on the marks obtained from your best TWO attempts.

Breakdown of marks for each question is shown in square parentheses.

All questions carry equal marks.

Students are permitted to use their own calculators subject to Faculty conditions.

Alpha-numeric memories must be cleared prior to the start of the examination.

S275 08/03/

  1. The circuit shown in Figure Q1 initially has no current and the switch is closed at time

t = 0.

Calculate:

(a) the currents i

1

and i

2

immediately after the switch is closed; [2]

(b) the maximum possible voltage across the capacitor; [3]

(c) the currents i

1

and i

2

when the voltage across the capacitor (v

c

) is at its maximum

value; [3]

(d) the time constant of the circuit whilst the switch is closed; [4]

(e) the time taken for the voltage across the capacitor to reach 10V. [5]

[Hint: for parts (d) and (e), use Thevenin’s theorem]

The switch is kept closed for a period of 1 second and then opened.

Calculate:

(f) the new time constant; [2]

(g) the time taken for the voltage across the capacitor to fall to 4V; [4]

(h) the magnitude and direction of the current i

2

at this time. [2]

You may assume that the equations relating the charging and discharging voltage (v

c

across a capacitor (C) in series with a resistor (R) are:

v V (1 e ) v V e

C o C o

= − =

− t −

RC

t

RC

and

where t is the elapsed time and V

o

is the final/initial capacitor voltage depending upon

whether the capacitor is charging or discharging.

t=

G

i

C

v

c

16V

2k Ω

μF

6k Ω

i

1

i

2

R

1

R

2

Figure Q

08/03/2004 continued

  1. An oscilloscope is used to display the waveforms of the two voltages as shown in Figure

Q3. It is known that there is no dc offset voltage for either of the two waveforms, and

that the lower trace is a full-wave rectified sinewave.

The settings of the oscilloscope are as follows:-

Y1 amplifier (upper trace) 1 V/div

Y2 amplifier (lower trace) 5 V/div

Timebase 0.5 ms/div

From the oscilloscope measurements and for each waveform, determine:

(a) the peak voltage; [2]

(b) the time period; [3]

(c) the frequency. [3]

In addition, confirm ONE average and ONE r.m.s. value of the waveforms given in the

following table:-

Y1 (Upper Trace) Y2 (Lower Trace)

(d) Average Voltage (V) 0.75 6.

(e) R.M.S. Voltage (V) 1.22 7.

[ 10 ]

[ 7 ]

Figure Q

END

S275 08/03/