Additional Answer Options: 9.2.2

2023 - May/June - Paper 1 - Question 9

9.1.1

North polecheck

light bulb Tips: 

  • Using Fleming's right-hand generator rule, when the force on side XY of the coil is upwards and the current is shown going from X to Y, the magnetic field must point away from Magnet pole A.
  • The magnetic field always points away from the north pole and towards the south pole, therefore, making A the north pole.

1 mark

9.1.2

 

Y to Xcheck

light bulb Tip: the diagram shows an alternating current generator which means that the direction of the current changes every 180°. Therefore, the current will flow from to X.

1 mark

9.1.3

checkcheckcheck 

light bulb Tips: 

  • The coil originally starts from a horizontal position which corresponds with the maximum (or minimum voltage).
  • Since it is an AC generator, the potential difference graph looks like a cos graph.
  • Two complete revolutions mean starting and ending at the same maximum with another maximum in the middle.

3 marks

(3 x 1 mark)

9.2.1

light bulb How to approach this question: 

  • Since the maximum current of 6A is given and the Vrms given is 200V, both values must be the rms or the max values to be able to calculate the resistance.

Option 1

V_{\text{rms }}=\frac{V_{\text{max }}}{\sqrt2}check

200=\frac{V_{\max}}{\sqrt2}check

V_{\text{max }}=282,84\mathrm{~V}

R=\frac{V}{I}

=\frac{282,84}{6}check

R=47,14\Omegacheck

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

4 marks

(4 x 1 mark)

9.2.2

Option 1

W=I^2R\Delta tcheck

=(4,24)^2(47,17)(7200)checkcheck

=6,11\times10^6\mathrm{J}\left(6,10\times10^6\right)check

light bulb Tips: 

  • The energy consumed by the device is given by the symbol W.
  • There are several different ways to calculate the energy consumed. You should use Vrms and Irms values regardless of which of the equations you select.

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

4 marks

(4 x 1 mark)


Or

Option 2

\mathrm{W}=\mathrm{VI}\Delta\mathrm{t}

=(200)(4,24)(7200)

=6,11\times10^6\mathrm{~J}


Or

Option 3

\mathrm{W}=\frac{\mathrm{V}^2\Delta\mathrm{t}}{\mathrm{R}}

=\frac{\left(200^2\right)(7200)}{47,27}

=6,11\times10^6\mathrm{J}\left(6,10\times10^6\right)


Or

Option 4

P=\frac{W}{\Delta t}

848=\frac{W}{7200}

=6,11\times10^6\mathrm{~J}

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