GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question1

Question2

Question3

Question4

Question5

GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question 1

In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains.

a) i) What is the name for this type of circuit? (1 mark)

ii) How much electrical energy does the 60W lamp convert in 5 minutes?

(3 marks)

iii) Not all the electrical energy is converted into light. What happens to the wasted energy?

(2 marks)

b) The lamps are now incorrectly connected as shown in the circuit below.

Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer.

(3 marks)

In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains.

a) i) What is the name for this type of circuit? (1 mark)

ii) How much electrical energy does the 60W lamp convert in 5 minutes?

(3 marks)

iii) Not all the electrical energy is converted into light. What happens to the wasted energy?

(2 marks)

Parallel circuit

Power is energy / time so energy = 60 x 5 = 300J

It is converted to heat energy

b) The lamps are now incorrectly connected as shown in the circuit below.

Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer.

(3 marks)

It will not give out as much energy because the first lamp is taking some of the electricity. They have to share the electricity

In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains.

a) i) What is the name for this type of circuit? (1 mark)

ii) How much electrical energy does the 60W lamp convert in 5 minutes?

(3 marks)

iii) Not all the electrical energy is converted into light. What happens to the wasted energy?

(2 marks)

Mark scheme

Parallel circuit

Power is energy per second, so the time must be converted into secondsEnergy = Power x time= 60 x 5 x 60 = 18kJ

It is converted to heat energy

The energy becomes spread out among many molecules and therefore irrecoverable, or “dissipated”.

b) The lamps are now incorrectly connected as shown in the circuit below.

Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer.

(3 marks)

Mark scheme

It will not give out as much energy.

The voltage across each lamp is now reduced and also the current through both lamps is reduced because the resistance has increased.

In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains.

a) i) What is the name for this type of circuit? (1 mark)

ii) How much electrical energy does the 60W lamp convert in 5 minutes?

(3 marks)

iii) Not all the electrical energy is converted into light. What happens to the wasted energy?

(2 marks)

Parallel circuit

Power is energy / time so energy = 60 x 5 = 300J

It is converted to heat energy

Power is energy per second, so the time must be converted into secondsEnergy = Power x time= 60 x 5 x 60 = 18kJ

Sufficient for 1 mark, but for the second mark needs a reference to the energy being spread out among many molecules and therefore irrecoverable, or “dissipated”.

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b) The lamps are now incorrectly connected as shown in the circuit below.

Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer.

(3 marks)

It will not give out as much energy because the first lamp is taking some of the electricity. They have to share the electricityJust scores the first mark, but does not show understanding that the

voltage across each lamp is now reduced and also the current through both lamps is reduced because the resistance has increased.

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GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question 2

A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second?

ii) In what form is the energy stored in the cell?

(2 marks)

(1 mark)

iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell?

(1 mark)

b) When the cell is connected across two identical lamps, each of resistance 2it delivers a current of 0.3A

What will be the rate of energy conversion in each lamp?

(3 marks)

A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second?

ii) In what form is the energy stored in the cell?

(2 marks)

(1 mark)

Energy = V x I = 1.5 x 0.2 = 0.3J

Electrical energy

iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell?

(1 mark)

b) When the cell is connected across two identical lamps, each of resistance 2it delivers a current of 0.3A

What will be the rate of energy conversion in each lamp?

(3 marks)

Heat

Energy = V x I = 1.5 x 0.3 = 4.5J

A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second?

ii) In what form is the energy stored in the cell?

(2 marks)

(1 mark)

Mark scheme

V x I = power (energy per second) 1.5 x 0.2 = 0.3W (0.3J per second)

Chemical energy

iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell?

(1 mark)

b) When the cell is connected across two identical lamps, each of resistance 2it delivers a current of 0.3A

What will be the rate of energy conversion in each lamp?

(3 marks)

Mark scheme

Thermal energy is dissipated in the cell.

I2R = 0.09 x 2 = 0.18J/s in each lamp.

A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second?

ii) In what form is the energy stored in the cell?

(2 marks)

(1 mark)

Energy = V x I = 1.5 x 0.2 = 0.3J

Electrical energy

V x I is actually power (energy per second) not total energy, but this is what the question asked for.

The cell converts from chemical energy to electrical energy.

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iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell?

(1 mark)

b) When the cell is connected across two identical lamps, each of resistance 2it delivers a current of 0.3A

What will be the rate of energy conversion in each lamp?

(3 marks)

Heat

Energy = V x I = 1.5 x 0.3 = 4.5J

Not an ideal answer, but adequate for 1 mark. Thermal energy is dissipated in the cell.

Scores 1 mark for V x I as rate of energy production, but the previous V is no-longer applicable. The cell is not necessarily giving 1.5V now, and each lamp has half the P.D. across it. V x I (with the appropriate, unknown, V) becomes I2R = 0.09 x 2 = 0.18J/s in each lamp.

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GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question 3

In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A.

a) i) How much energy is converted in the lamp in 2 minutes?

(3 marks)

ii) Calculate the resistance of the motor.

(3 marks)

b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not.

(3 marks)

In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A.

a) i) How much energy is converted in the lamp in 2 minutes?

(3 marks)

Energy = 4.5 x 0.25 x 2 = 2.25J

ii) Calculate the resistance of the motor.

(3 marks)

b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not.

(3 marks)

V = I x R R = 4.5/0.25 = 18 Ohms

The motor is stronger, it would take all the energy. The lamp needs 0.25 not 0.1 like the motor.

In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A.

a) i) How much energy is converted in the lamp in 2 minutes?

(3 marks)

Mark scheme

Energy = 4.5 x 0.25 x 2 x 60 = 135J

ii) Calculate the resistance of the motor.

(3 marks)

b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not.

(3 marks)

Mark scheme

V = IR R = 4.5/0.1 = 45

The motor has higher resistance, therefore would have a greater proportion of the P.D. across it. The combined resistance would lower the current, leaving not enough power (V x I) to light the lamp.

In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A.

a) i) How much energy is converted in the lamp in 2 minutes?

(3 marks)

Energy = 4.5 x 0.25 x 2 = 2.25JScores for method, but time should be in seconds.

Energy = 4.5 x 0.25 x 2 x 60 = 135J

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ii) Calculate the resistance of the motor.

(3 marks)

b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not.

(3 marks)

V = I x R R = 4.5/0.25 = 18 ohms

The motor is stronger, it would take all the energy. The lamp needs 0.25 not 0.1 like the motor.

Method is correct, but the wrong current has been used. The current in the motor branch is 0.1A. R = 4.5/0.1 = 45

Stronger is not an appropriate word. The motor has higher resistance, therefore would have a greater proportion of the P.D. across it. The combined resistance would lower the current, leaving not enough power (V x I) to light the lamp.

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GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question 4

A solar cell is used to power a small electric motor.

a) This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V.

i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell.

(3 marks)

ii) Calculate the rate, in J/s, at which energy is supplied to the motor.

(1 mark)

iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why.

(2 marks)

b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this?

(2 marks)

A solar cell is used to power a small electric motor.

a) This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V.

i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell.

(3 marks)

Energy = VIt = 12/3 x 200 = 0.02

ii) Calculate the rate, in J/s, at which energy is supplied to the motor.

(1 mark)

iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why.

(2 marks)

b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this?

(2 marks)

Power = V x I = 3 x 0.02 = 0.06 J/s

Energy is lost as heat

One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000

A solar cell is used to power a small electric motor.

a) This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V.

i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell.

(3 marks)

Mark scheme

Energy = VIt I = Energy/Vxt I = 12/(3 x 200) = 0.02A

ii) Calculate the rate, in J/s, at which energy is supplied to the motor.

(1 mark)

iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why.

(2 marks)

b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this?

(2 marks)

Mark scheme

Power = V x I 3 x 0.02 = 0.06 J/s

Energy is wasted as heat. Low efficiency.

One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000

A solar cell is used to power a small electric motor.

a) This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V.

i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell.

(3 marks)

Energy = VIt = 12/3 x 200 = 0.02

Although the calculation is badly expressed, marks have been awarded for a correct calculation; however, if the calculation had been only partly correct, the way it is expressed could have caused loss of a mark which would have otherwise been awarded. Units have been omitted

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ii) Calculate the rate, in J/s, at which energy is supplied to the motor.

(1 mark)

iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why.

(2 marks)

b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this?

(2 marks)

Power = V x I = 3 x 0.02 = 0.06 J/s

Energy is lost as heat

One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000

The answer could also be obtained from 12J of energy in 200s, so energy per s = 12J/200s = 0.06J/s

Scores for heat, but energy is not “lost”; it is wasted. Reference to efficiency could also score.

45

GCSE Physics Exam Doctor

Electricity – Energy in Circuits

Question 5

A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done?

(2 marks)

ii) How much gravitational potential energy does the load gain?

(1 mark)

b) i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second?

(2 marks)

ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether?

(2 marks)

(1 mark)

iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor

A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done?

(2 marks)

ii) How much gravitational potential energy does the load gain?

(1 mark)

b) i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second?

(2 marks)

Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm

0.5 J

6 X 0.1 = 0.6

ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether?

(2 marks)

(1 mark)

iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor

0.6 x 4 = 2.4J

0.5/2.4 = 0.21%

A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done?

(2 marks)

ii) How much gravitational potential energy does the load gain?

(1 mark)

b) i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second?

(2 marks)

Mark scheme

Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm

Energy gained = work done. e.c.f from part i) would be allowed.Units can be Nm or J for either part.

0.5 Nm

6 X 0.1 = 0.6J

ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether?

(2 marks)

(1 mark)

iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor

Mark scheme

0.6 x 4 = 2.4J

0.5/2.4 = 0.21 or 21%

A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done?

(2 marks)

ii) How much gravitational potential energy does the load gain?

(1 mark)

b) i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second?

(2 marks)

Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm

0.5 J

6 X 0.1 = 0.6

Energy gained = work done. e.c.f from part i) would be allowed.Units can be Nm or J for either part.

Units are omitted.

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ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether?

(2 marks)

(1 mark)

iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor

0.6 x 4 = 2.4J

0.5/2.4 = 0.21%Efficiency can be expressed as a fraction e.g. 0.21 or as a percentage 21%,

but it is not 0.21%

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GCSE Physics Exam Doctor

Electricity – Energy in Circuits

End of questions