• Physics 9702 Doubts | Help Page 17

Question 90: [Current of Electricity > Resistance]
Battery of electromotive force (e.m.f.) V and negligible internal resistance is connected to 1 kΩ resistor, as shown.

Student attempts to measure potential difference (p.d.) between points P and Q using two voltmeters, one at a time. First voltmeter has a resistance of 1 kΩ and the second voltmeter has a resistance of 1 MΩ.
What are the readings of the voltmeters?

Reference: Past Exam Paper – June 2014 Paper 12 Q30



Solution 90:
Answer: B.
The voltmeter is connected in series with the resistor. So, the potential difference across a component R₁is given by
V₁ = [R₁ / (R₁ + R₂)] V           (potential divider equation)
where R₁ is voltmeter resistance here, R₂ = 1kΩ (resistance of resistor) and V is the e.m.f.

For the voltmeter with 1kΩ (R₁ = R₂) resistance,
V₁ = [R₂ / (R₂ + R₂)] V = [R₂ / 2R₂] V = V /2

For the 1MΩ resistance (R₁ >> R₂) voltmeter, ([R₁ + R₂] ≈ R₁ since R₁ >> R₂)
V₁ = [R₁ / (R₁ + R₂)] V ≈ [R₁ / R₁] V = V

 






Question 91: [Current of Electricity > Current]
Battery of negligible internal resistance is connected to a resistor network, ammeter and switch S, as shown.

When S is open, reading on ammeter is 250 mA.
When S is closed, what is the change in reading on ammeter?
A 1.07 A                     B 1.32 A                     C 190 mA                   D 440 mA

Reference: Past Exam Paper – June 2014 Paper 13 Q38



Solution 91:
Answer: C.
The ammeter reads the total current in the circuit.

When switch S is open, the 2.8Ω resistor is short-circuited.
Total resistance in circuit = 4.8 + 7.2 = 12.0Ω
e.m.f of battery = IR = (250x10^-3) x 12 = 3V

When switch S is closed, all resistors need to be considered.
Total resistance in circuit = 4.8 + [(1/7.2) + (1/2.8)]^-1 = 6.816Ω
Total current in circuit (= e.m.f. / R) = 3 / (6.816) = 0.440A = 440mA

Change in ammeter reading = 440 – 250 = 190mA









Question 92: [Current of Electricity > Current]
Diagram shows a four-terminal box connected to battery and two ammeters.

Currents in the two meters are identical.
Which circuit, within box, will give this result?

Reference: Past Exam Paper – November 2012 Paper 13 Q35



Solution 92:
Go to
The diagram shows a four-terminal box connected to a battery and two ammeters.









Question 93: [Power > Efficiency > Wind Turbine]
Wind turbine has blades that sweep an area of 2000 m². It converts power available in wind to electrical power with efficiency of 50%.
What is the electrical power generated if wind speed is 10 m s^–1? (Density of air is 1.3 kg m^–3.)
A 130 kW                   B 650 kW                    C 1300 kW                  D 2600 kW

Reference: Past Exam Paper – June 2013 Paper 11 Q18



Solution 93:

Answer: B.

The kinetic energy of the air causes the wind turbine to rotate which is then converted to electrical power.

Kinetic energy of air = ½ mv². The speed is known (= 10ms^-1), so we need to find the mass.

Density = mass / volume

Density of air = 1.3kgm^-3

The wind speed is 10ms^-1. This wind blows on an area of 2000m². This may be interpreted as follows: In one second, a column of air 10m length and with an area of 2000m² moves past the blades of the wind turbine. This is equivalent to a volume of (10 x 2000 =) 20 000m³ passing the blades per second.

Mass = Density x Volume. Since we have the volume of air per second,

Mass of air passing the blades per second = Density x Volume of air per second

Mass of air passing the blades per second = 1.3 x 20 000 = 26 000 kgs^-1

Kinetic energy = ½ mv²

Power from wind = Energy / time = (½ mv²) / t = ½ (m/t) v²

where (m/t) is the mass of air passing the blades per second  

Efficiency = 50%. That is, only 50% of the power of the wind is converted to electrical power.

Electrical Power = 0.50 x [0.5 x 26000 x 10²] = 650 000W = 650kW

Question 94: [Forces > Resultant forces]

A lift (elevator) consists of passenger car supported by cable which runs over a light, frictionless pulley to balancing weight. Balancing weight falls as passenger car rises.

Some masses are shown.

What is the magnitude of acceleration of car when carrying just one passenger and when pulley is free to rotate?

A 0.032 m s^–2              B 0.32 m s^–2                C 0.61 m s^–2                D 0.65 m s^–2

Reference: Past Exam Paper – June 2013 Paper 13 Q9

Solution 94:

Answer: B.

The passenger car (mass = 520), balancing weight (mass = 640) and passenger (mass = 80) will have a weight of 520g, 640g and 80g respectively, where g is the acceleration due to gravity.

The weight of the passenger car and passenger causes to car to go downwards, while the weight of the lift causes the car to move upwards (the forces they act on the car are in opposite directions).

Resultant (upward) force on car = 640g – (520g + 80g) = 40g

Acceleration due to gravity = 9.81ms^-2

Resultant force on car = 40 x 9.81 = 392N

This resultant (unbalanced) force of 392 N has to accelerate all the mass, (640 + 520 + 80 =) 1240 kg, and not just the mass of the lift and passenger.

ma = 392

Acceleration, a = 392 / 1240 = 0.32ms^-2