The display on a cathode-ray oscilloscope shows the signal produced by an electronic circuit.

Question 19

The display on a cathode-ray oscilloscope shows the signal produced by an electronic circuit.

The time-base is set at 5.0 ns per division and the Y-gain at 10 V per division.

What is the frequency of the signal?

A 2.0 × 10^-8 Hz

B 2.5 × 10^-2 Hz

C 5.0 × 10⁷ Hz

D 3.1 × 10⁸ Hz

Reference: Past Exam Paper – June 2014 Paper 13 Q5

Solution:

Answer: C. 

Considering the 2 peaks, it can be observed that a complete wave occupies 4 divisions (horizontally).

Time-base = 5.0 ns / div

1 division - - > 5.0 ns

4 divisions - - > 4 × 5.0 = 20 ns

So, the period T = 20 ns

 
Frequency = 1 / T = 1 / (20×10^-9) = 5.0×10⁷ Hz

Two parallel plates R and S are 2 mm apart in a vacuum. An electron with charge –1.6 × 10-19 C moves along a straight line in the electric field between the plates.

Question 17

Two parallel plates R and S are 2 mm apart in a vacuum. An electron with charge –1.6 × 10^-19 C moves along a straight line in the electric field between the plates. The graph shows how the potential energy of the electron varies with its distance from plate R.

Which deduction is not correct?

A The electric field between R and S is uniform.

B The electric field strength is 3000 N C^-1.

C The force on the electron is constant.

D The magnitude of the potential difference between R and S is 3 V.

Reference: Past Exam Paper – November 2015 Paper 13 Q29

Solution:

Answer: B.

The electric field between two parallel plates is uniform.

Electric field strength, E = F / q

Force on the charge: F = Eq

Since the electric field strength is constant, the force on the electron is also constant.

(Potential energy) Work done = Force × distance, s

Work done = (Eq) × s

From the graph, when distance s = 2 mm, potential energy (work done) = 4.8×10^-19 J

Work done = (Eq) × s

4.8×10^-19 = E × 1.6×10^-19 × 2×10^-3

Electric field strength, E = (4.8×10^-19) / (1.6×10^-19 × 2×10^-3) = 1500 N C^-1

So, the electric field strength E is 1500 N C^-1, not 3000 N C^-1. Deduction B is NOT correct. [Choice B is the answer]

For parallel plates,

E = V / d

Potential difference, V = Ed = 1500 × 2×10^-3 = 3 V

A station on Earth transmits a signal of initial power 3.1 kW to a geostationary satellite. The attenuation of the signal received by the satellite is 190 dB.

Question 7

(a) Information may be carried by different channels of communication.

State one application, in each case, where information is carried using

(i) microwaves, [1]

(ii) coaxial cables, [1]

(iii) wire pairs. [1]

(b) A station on Earth transmits a signal of initial power 3.1 kW to a geostationary satellite.

The attenuation of the signal received by the satellite is 190 dB.

(i) Calculate the power of the signal received by the satellite. [2]

(ii) By reference to your answer in (i), state and explain the changes made to the signal

before transmission back to Earth. [3]

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q12

Solution:

(a)

(i) e.g. satellite communication, mobile phones, line of sight communication, wifi

(ii) e.g. connection of TV to aerial, loudspeaker, microphone (if clearly identified)

(iii) e.g. a.f. amplifier to loudspeaker, landline for phone

(b)

(i)

Attenuation (in dB) = 10 lg (P_IN / P_OUT)

P_OUT is the power received by the satellite and P_IN is the initial power of the signal.

190 = 10 lg (3.1×10³ / P_OUT)

lg (3.1×10³ / P_OUT) = 190 / 10

3.1×10³ / P_OUT = 10¹⁹

P_OUT = 3.1×10³ /10¹⁹ = 3.1×10^-16 W = 3.1×10^-19 kW

(ii) The signal is amplified and its frequency is changed to prevent swamping of the up-link signal by the down-link (signal).