Question 9.2.3: Additional Answer Options

2020 - November - Paper 1 - Question 9

9.1.1

 

DC motor check

light bulb Tip: the presence of a cell/power source and split rings indicate that the machine is a DC motor.

1 mark

 

9.1.2

 

Electrical to mechanical/kinetic energy checkcheck

light bulb Tip: a motor converts electrical energy into mechanical energy, specifically kinetic energy.

2 marks

(1 x 2 marks)

9.1.3

 

Split ring/commutator check

1 mark

 

9.1.4

 

Anticlockwise checkcheck

light bulb Tips: 

  • The current flows around the circuit in the direction shown in the image below.
  • Using Fhleming's left-hand motor rule, the direction of the force can be determined. This shows that the coil rotates in an anticlockwise direction.

 

2 marks

(1 x 2 marks)

9.2.1

 

(The RMS voltage/value of AC is) the AC voltage/potential difference which dissipates the same amount of energy/heat/power as an equivalent DC voltage/potential difference.checkcheck

Or

The RMS voltage/value of AC is the DC potential difference which dissipates the same amount of energy/heat/power as AC. checkcheck

2 marks

(1 x 2 marks)

9.2.2

 

Option 1

P_{\text{ave }}=\frac{V_{ms}^2}{R}check

200=\frac{220^2}{R}check

R=242\Omegacheck

Or

Option 2

P_{\mathrm{ave}}=\mathrm{V}_{\mathrm{rms}}\mathrm{I}_{\mathrm{ms}}check

200=I_{\mathrm{rms}}(220)

I_{\mathrm{rms}}=0,909\mathrm{~A}(0,91)

\mathrm{R}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{ms}}}\text{ OR }\mathrm{R}=\frac{\mathrm{V}}{\mathrm{l}}

R=\frac{220}{0,909}check

\mathrm{R}=242\Omega(241,76\Omega)check

Or

Option 3

P_{\text{ave }}=V_{\text{ms }}I_{ms}check

200=I_{\mathrm{rms}}(220)

I_{\mathrm{rms}}=0,909\mathrm{~A}(0,91)

P_{\text{ave }}=I_{rms}{}^2R

200=(0,909)^2Rcheck

R=242\Omegacheck

3 marks

(3 x 1 mark)

9.2.3

 

Option 1

\mathrm{W}=\frac{\mathrm{V}^2\Delta\mathrm{t}}{\mathrm{R}} (1 + 2) checkcheck

=\frac{\left(150^2\right)(10\times60)}{242}(3 + 4)checkcheck

=55785,12\mathrm{~J} (5) check

light bulb Tip: the time must be in seconds, therefore, minutes are converted to seconds.

There are more ways to answer this question. To view other options, click here.

5 marks

(5 x 1 mark)

Option 1 marking criteria:

  • (1 + 2) Appropriate formula for W containing Vcheckcheck
  • (3 + 4) Substitution checkcheck
  • (5) Final answer: 55785,12 J check

Marking criteria:

  • (1) Appropriate formula to calculate P or Irms.
  • (2) Substitution.
  • (3) Formula for P or W containing Δt.
  • (4) Substitution.
  • (5) Final answer: 55 785,12 J.

Or

Option 2

P_{ave}=\frac{V_{ms}^2}{R} (1)

=\frac{150^2}{242} (2)

P_{av}=92,975\mathrm{~W}

P=\frac{W}{\Delta t} (3)

92,975=\frac{W}{(10)(60)} (4)

W=55785,12\mathrm{~J} (5)

(55785,12-55896\mathrm{~J})


Or

Option 3

R=\frac{V_{ms}}{I_{ms}}/R=\frac{V}{I} (1)

242=\frac{150}{I_{\mathrm{ms}}} (2)

I_{\mathrm{ms}}=0,620\mathrm{~A}

P_{\text{ave }}=I_{\text{ms }}V_{\text{rms }}

=(0,62)(150)

=92,97\mathrm{~W}(93\mathrm{~W})

P=\frac{W}{\Delta t} (3)

92,975=\frac{W}{(10)(60)} (4)

W=55785,12\mathrm{~J} (5)

(55785,12-55896\mathrm{~J})


Or

Option 4

Part 1

R=\frac{V_{\text{ms }}}{I_{\text{rms }}}/R=\frac{V}{I} (1)

242=\frac{150}{I_{\mathrm{rms}}} (2)

I_{\mathrm{rms}}=0,620\mathrm{~A}

Part 2

\mathrm{W}=\mathrm{I}^2\mathrm{R}\Delta\mathrm{t} (3)

=(0,62)^2(242)(10)(60) (4)

=55814,88\mathrm{~J} (5)

(55785,12-55896\mathrm{~J})

Or

\mathrm{W}=\mathrm{VI}\Delta\mathrm{t} (3)

=(150)(0,62)(600) (4)

=55800\mathrm{~J} (5)


Or

Option 5

P_{\text{ave }}=\frac{V_{ms}^2}{R}=\frac{150^2}{242}=92,975\mathrm{W}

P_{\text{ave }}=I_{rms}{}^2R (1)

92,975=I_{\mathrm{rms}^2}^2(242) (2)

\mathrm{I}_{\mathrm{rms}}=0,6198\mathrm{~A}

\mathrm{W}=\mathrm{I}^2\mathrm{R}\Delta\mathrm{t} (3)

=(0,6198)^2(242)(10)(60) (4)

=55778,88\mathrm{~J} (5)

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