• Physics 9702 Doubts | Help Page 241

Question 1118: [Current of Electricity > Potential divider]

A potential divider circuit consists of fixed resistors of resistance 2.0 Ω and 4.0 Ω connected in series with a 3.0 Ω resistor fitted with a sliding contact. These are connected across a battery of e.m.f. 9.0 V and zero internal resistance, as shown.

What are the maximum and the minimum output voltages of this potential divider circuit?

maximum voltage / V              minimum voltage / V

A                     4.0                                           2.0

B                     5.0                                           2.0

C                     9.0                                           0

D                     9.0                                           2.0

Reference: Past Exam Paper – June 2015 Paper 12 Q36

Solution 1118:

Answer: B.

The potential at the top of the circuit (connected to the + terminal) is 9 V and the potential at the bottom of the circuit (connected to the – terminal) is 0 V.

There will be a p.d. across each of the fixed resistors.

The output voltage is maximum when the maximum value of the variable resistor is taken (= 3.0 Ω). [We need to consider the total resistance across the terminals.]

Max. output voltage = [(2+3) / (2+3+4)] × 9.0 = 5.0 V

The output voltage is minimum when the minimum value of the variable resistor is taken (= 0 Ω).

Min. output voltage = [(2+0) / (2+3+4)] × 9.0 = 2.0 V

Question 1119: [Electromagnetism]

A uniform magnetic field of flux density B makes an angle θ with a flat plane PQRS, as shown in Fig. 5.1.

Fig. 5.1

The plane PQRS has area A.

(a) State

(i) what is meant by a magnetic field,

(ii) an expression, in terms of A, B and θ, for the magnetic flux Φ through the plane PQRS.

(b) A vertical aluminium window frame DEFG has width 52 cm and length 95 cm, as shown in Fig. 5.2.

Fig. 5.2

The frame is hinged along the vertical edge DG.

The horizontal component B_H of the Earth’s magnetic field is 1.8 × 10^–5 T. For the closed window, the frame is normal to the horizontal component B_H.

The window is opened so that the plane of the window rotates through 90°.

(i) Explain why, when the window is opened, the change in magnetic flux linkage due to the vertical component of the Earth’s magnetic field is zero.

(ii) Calculate, for the window opening through an angle of 90°, the change in magnetic flux linkage.

(c) (i) State Faraday’s law of electromagnetic induction.

(ii) The window in (b) is opened in a time of 0.30 s.

Use your answer in (b)(ii) to calculate the average e.m.f. induced in the window frame.

(iii) State the sides of the window frame between which the e.m.f. is induced.

Reference: Past Exam Paper – November 2013 Paper 43 Q5

Solution 1119:

(a)

(i) A magnetic field is a region (of space)                   EITHER where a moving charge (may) experience a force                        OR around a magnet where another magnet experiences a force

(ii) (Φ =) BA sinθ

{We need to consider the perpendicular component to the plane.}

(b)

(i) The plane of frame is always parallel to B_V (vertical component of the Earth’s magnetic field) / flux linkage always zero

(ii)

{Φ = BA sin90 = BA. We need to convert the area to SI unit.}

ΔΦ = 1.8×10^–5 × (52×10^–2 × 95×10^–2)

ΔΦ = 8.9 × 10^–6 Wb

(c) (i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage).

(ii)

{From the above definition, e.m.f. = ΔΦ / t}

e.m.f. = (8.9 × 10^–6) / 0.30 = 3.0 × 10^–5 V

(iii) This question part was removed from the assessment. All candidates were awarded 1 mark. 

Question 1120: [Stationary waves]

A musical instrument called a bugle is a long tube with a mouthpiece at one end. The other end is open and flared, as shown.

A musician maintains stationary sound waves with a node at the mouthpiece and an antinode at the other end. The lowest frequency of sound that the bugle can produce is 92 Hz.

Which different frequencies of sound can be produced by the bugle?

A 92 Hz, 138 Hz, 184 Hz, 230 Hz, 276 Hz

B 92 Hz, 184 Hz, 276 Hz, 368 Hz, 460 Hz

C 92 Hz, 276 Hz, 460 Hz, 644 Hz, 828 Hz

D 92 Hz, 276 Hz, 828 Hz, 2484 Hz, 7452 Hz

Reference: Past Exam Paper – March 2016 Paper 12 Q25

Solution 1120:

Answer: C.

A node (amplitude = 0) is found at the mouthpiece and an antinode (maximum amplitude) is found at the other end. This holds for all the frequencies that can be produced.

The lowest frequency occurs when the length of the instrument is just a quarter of a wavelength between a node and an antinode.

¼ λ corresponds to a frequency of 92 Hz.

The next frequency possible is when the instrument length is three-quarters of a wavelength.

¾ λ would correspond to 3 × 92 = 276 Hz

The next frequency possible is when the instrument length is 5/4 of a wavelength.

5λ/4 would correspond to 5 × 92 = 460 Hz

Next is when the length is 7λ /4 and the following is when the length is 9λ/4 ….