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Higher Certificate in Engineering in Electronic Engineering – Award
(NFQ Level 6)
Summer 2006
ELECTROTECHNOLOGY
(Time: 3 Hours)
Answer any 5 questions All questions carry equal marks.
Examiners: Dr. J. Connell Dr. R O Dubhghaill Mr. J. Berry
Q1.
For the circuit in Figure 1, calculate:
(i) The main circuit current I 1 (8 marks) (ii) The voltage at node A (4 marks) (iii) The current in L1 (4 marks) (iv) The voltage at node B (4 marks)
Q2.
(a) State both Thevenin’s and Norton’s theorems (4 marks) (b) What is the relationship between the Thevenin open circuit voltage and the Norton short circuit current? (2 marks) (c) For the circuit below, use Norton’s theorem to calculate the load current if the load is an impedance of (4+j6)Ω. (14 marks)
VAC
5V,
1MHz
3.3nF 10 μH
A B
I 1
Q3.
(a) Draw a plot of the current versus frequency in a series resonant circuit. (2 marks) (b) Draw a plot of input impedance versus frequency in a parallel resonant circuit. (2 marks) (c) In a series resonant circuit, why do the inductor and capacitor seem to “disappear” at resonance? (2 marks) (d) Why is a series resonant circuit potentially dangerous? (2 marks) (e) A parallel resonant circuit consists of a coil, having an inductance of 82μH and a winding resistance of 18Ω, in parallel with a variable capacitor. The resonant circuit is connected to a supply having a frequency of 500kHz. Calculate:
(i) The value of C to give resonance. (3 marks) (ii) The dynamic impedance and the Q-factor of the circuit (3 marks) (iii) The bandwidth of the circuit (3 marks) (iv) The upper and lower cut off frequencies. (3 marks)
Q4.
(a) Briefly explain the differences between an attenuator and a filter (3 marks)
(b) For an RL high pass filter with L=27mH and R=2.2kΩ
(i) the cutoff frequency (2 marks) (ii) the attenuation at 5kHz (5 marks) (iii) the attenuation at 50Hz (3 marks) (iv) the rolloff rate (2 marks)
(c) If the input frequency in (b) is 1kHz and the input amplitude is 2V, calculate the amplitude of the output voltage. (4 marks)
( 100 + j 0 ) V
R = 6 Ω XL = 8 Ω
X (^) C = 3 Ω Load
