The diagram shows an electric field pattern caused by two positive and two negative point charges of equal magnitude placed at the four corners of a square.

Question 9

The diagram shows an electric field pattern caused by two positive and two negative point charges of equal magnitude placed at the four corners of a square.

In which direction does the force act on an electron at point X?

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q31

Solution:

Answer: B.

The direction of the electric field is from positive to negative.

The electric field lines at any point show the direction of the force on a positive test charge. Since we are dealing with an electron (which is has a negative charge), the direction of the force on it would be opposite to the direction of the force on a positive charge (which is indicated by the direction of the field).

But when the field lines are curved, the direction of the force is along the tangent at that point. [A and C are incorrect] The tangent is parallel to the field lines.

D shows the direction of the force on a positive charge as this is in the same direction as the field at that point.

The direction of the force on an electron would be opposite, which is along B.

Fig. 10.1 shows the variation with frequency f of the power P of a radio signal.

Question 3

Fig. 10.1 shows the variation with frequency f of the power P of a radio signal.

Fig. 10.1

(a) State the name of

(i) the type of modulation of this radio signal, [1]

(ii) the component of frequency 50 kHz, [1]

(iii) the components of frequencies 45 kHz and 55 kHz. [1]

(b) State the bandwidth of the radio signal. [1]

(c) On the axes of Fig. 10.2, sketch a graph to show the variation with time t of the signal

voltage of Fig. 10.1. [3]

Reference: Past Exam Paper – November 2007 Paper 4 Q10

Solution:

(a)

(i) Amplitude (modulated)     

(ii) Carrier (frequency / wave)

(iii) Sideband (frequency)

(b) Bandwidth = 10kHz        

(c)

The sketch should have a general shape, i.e. any wave that is amplitude modulated           

with a correct period for the modulating waveform (200μs)             {corresponding to frequency 5 kHz}

and a correct period for the carrier waveform (20μs)             {corresponding to frequency 50kHz  

A simple circuit is formed by connecting a resistor of resistance R between the terminals of a battery of electromotive force (e.m.f.) 9.0 V and constant internal resistance r.

Question 21

A simple circuit is formed by connecting a resistor of resistance R between the terminals of a battery of electromotive force (e.m.f.) 9.0 V and constant internal resistance r.

A charge of 6.0 C flows through the resistor in a time of 2.0 minutes causing it to dissipate 48 J of thermal energy.

What is the internal resistance r of the battery?

A 0.17 Ω          B 0.33 Ω          C 20 Ω            D 160 Ω

Reference: Past Exam Paper – June 2015 Paper 13 Q34

Solution:

Answer: C.

A charge of 6.0 C flows through the resistor in a time of 2.0 minutes.

Current I = Q / t

Current through resistor = 6.0 / (2×60) = 0.05 A

Power dissipated in a resistor = I²R

Power = Energy / time

So, Energy dissipated = P × t = I²Rt = 48 J

(0.05)² × R × (2×60) = 48

Resistance R = 48 / [(0.05)² × (2×60)] = 160 Ω

The internal resistance of the battery and the resistor are in series. The same current flows through them.

So, the sum of p.d. across these 2 is equal to the e.m.f. in the circuit.

I × (R + r) = 9.0

R + r = 9.0 / 0.05 = 180

160 + r = 180

Internal resistance r = 180 – 160 = 20 Ω