Thursday, November 20, 2014

Physics 9702 Doubts | Help Page 17

  • 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:
Answer: D.
The ammeter connected close to the battery gives the total current in the circuit while the ammeter connected between terminals 3 and 4 give the current passing through terminal 3 to 4 (current in this one will only be equal to the one close to the battery only if all components are connected in series in the whole circuit, including within the box).

For circuit A, the section from terminal 2 to 4 and finally to 3 is short-circuited (since 1 and 3 are not connected). So, no current flows through the ammeter.

For circuit B, the ammeter connected to 3 and 4 is short-circuited since wires (which have negligible resistance) are connected between 1 to 3, 3 to 4 and 4 to 2.

For circuit C, a parallel connection is formed between the 2 resistors between 1 to 3 and 1 to 2. So, the current splits up at junction 1. The second ammeter is connected in series with one of the resistors, so it only gives the current flowing in one resistance.

For circuit D, the 2 ammeters are in series with the resistor between terminals 1 and 3. Current is constant in for components connected in series.

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


  1. in q93, if we use the formula P=Fv, to calculate F, will the acceleration be g=9.81 ms^-2?

    1. I don't think this formula can be used here.
      Try to show your full working if you worked it that way.

  2. this is how i did it:
    =density*volume*acc. due to gravity*velocity*50%
    this gives an incorrect in case we use P=Fv, what should be the acc. used to calculate F?
    why does air not fall to the ground?as in, shouldn't all the air particles be present at the surface of the earth due to gravity and because of their negligible mass?

    1. Acceleration due to gravity acts DOWNWARDS while the air are passing the blades horizontally. The method you used in incorrect. The equation P = Fv cannot be used here. You should do it as I explained.

      The force of gravity acting on a body depends on the mass of the body. This force is also called weight. If the mass is negligible (as for air particles), the weight is also negligible.

  3. for quesion 90 here I dont get how you got V for the 1 MΩ resistor

    1. Since R1 which is 1MΩ is much larger than R2 which is 1kΩ,
      R1 + R2 = 1MΩ + 1kΩ is approximately equal to 1MΩ
      that is, R1 + R1 is approximately equal to R1

      1 000 000 + 1 000 = 1 001 000. this is approximately equal to 1 000 000. there is not much difference.

  4. Hi,
    Actually I have doubts in the following questions could you please help,

    Thanks a lot,

    1. I checked them. Most of them are already solved here. You need to search for the corresponding years at

      Alternatively, go to google search an type

      followed by the first few lines of the question.

  5. In q 93 can i use formula p=fv and i find force through rate of change of momentum assuming intial velocity zero.. And i am a bit confused related to this formula .. Can you please calirfy that the fromula p=fv can i use it when a object is accelrating or this only used when accelration is zero

    1. Only when velocity is constant, that is, acceleration is zero.

      I don't think you can use this formula here. Did you obtain the same answer. If yes, try to write your workings here.

  6. in question 90 i don't get how do you know that the voltmeter is connected in series and not in parallel?

    1. That's the basic the circuits. Voltmeters are connected across a component to read the p.d. across it while ammeters are connected in series to read the current flowing through it.

      However here, the voltmeter is a measuring the p.d. across P and Q.

      To know whether 2 components are connected in series or parallel, follow the flow of currents.
      Current starts from the +ve terminal of the battery, through the resistor, then through the voltmeter and finally back to the -ve terminal of the batttery.
      The same current flows through the resistor and the voltmeter - so they are in series.

      If the current was split up before going into the resistor and the voltmeter, then the 2 would have been in parallel.

  7. Wish I had found this blog a few weeks ago.. My exam is tomorrow and I'm not ready. Will definitely revisit next year when I do the complete A Level exam and refer to my classmates. Keep up the great posts. Cheers!

  8. The formula used is correct but there are also other ways to solve it as every problems has different solutions depending on the person solving it.

    ammeter 101

  9. Replies
    1. see solution 1110 at

  10. In Q91 I dont get why you used the power of -1 while calculating the Total Resistance when Switch is CLOSED. Can you please Explain.

    1. when S is closed, the 2.8 and 7.2 resistors are now in parallel. so we use the formula for parallel resistors


If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 17