Additional Answer Options: 10.2.2

2018 - May/June - Paper 1 - Question 10

10.1.1

Split ring/commutator check

light bulb Tip: a DC motor contains a split ring or commutator. The split ring is also shown clearly in the diagram below:

1 mark

10.1.2

Anti-clockwise checkcheck

light bulb Tip: using Fleming's left-hand motor rule, you can determine that the force acts downwards and this causes the coil to rotate in an anticlockwise direction.

2 marks 

(1 x 2 marks)

10.1.3

Electrical energycheck to mechanical(kinetic) energy. check 

light bulb Tip: a motor converts electrical energy provided by the cell into mechanical energy.

2 marks

(2 x 1 mark)

10.2.1

DC generator: split ring/commutator and AC generator have slip rings. check 

Or

AC generator: the slip ring and DC generator have split rings. check 

light bulb Tips: 

  • AC and DC generators have different connections to the external circuit.
  • A DC generator has a split ring or a commutator, and an AC generator has slip rings.

1 mark

10.2.2

\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{\max }}{\sqrt{2}} check 

=\frac{320}{\sqrt{2}} check 

=226,27 \mathrm{~V} check 

light bulb Tip: remember that root means the square voltage is the maximum voltage divided by \sqrt2.

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

3 marks

(3 x 1 mark)

10.2.3

Option 1

I_{\max }=\frac{V_{\max }}{R} 

=\frac{320}{35} check 

=9,14 \mathrm{~A} 

light bulb Tips: 

  • This formula is not on the formula sheet, however, it is an exceptionally useful equation to use in exams.
  • As long as the current and the potential difference are both written as maximum or the rms value, the equation is accepted.

\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\max }}{\sqrt{2}} check 

=\frac{9,14}{\sqrt{2}} check 

=6,46 \mathrm{~A} check 

Or

Option 2

Part 1:

P_{\text {average }}=\frac{V_{\text {rms }}^2}{R} 

=\frac{226,27^2}{35} check 

=1462,80 \mathrm{~W} 

Part 2:

P_{\text {average }}=V_{\text {rms }} I_{m s} check 

1462,80=(226,27)_{\mathrm{rms}} check 

I_{r m s}=6,46 A check 

Or

P_{\text{average }}=I_{\mathrm{rms}}^2R check 

1462,80=1_{\mathrm{ms}}^2(35) check 

\mathrm{I}_{\mathrm{rms}}=6,46 \mathrm{~A} check 

Or

Option 3

I_{r m s}=\frac{V_{r m s}}{R} check 

=\frac{226,27}{35} check check 

=6,46 \mathrm{~A} check 

4 marks

(4 x 1 mark)


Or

Option 2

\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{max}}}{\sqrt2}

V_{\max}=(240)\sqrt2

=339,41

V_{\max}=I_{\max}R

I_{\max}=\frac{339,41}{11}

I_{\max}=30,86\mathrm{~A}


Or

Option 3

Part 1

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

P_{\text{ave }}=\frac{(240)^2}{11}

P_{\text{ave }}=5236,36\mathrm{~W}

Part 2

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

5236,36=I_{rms}240

I_{rms\text{ }}=21,82\mathrm{A}

I_{rms}=\frac{I_{\max}}{\sqrt2}

21,82=\frac{I_{\max}}{\sqrt2}

I_{\max}=30,86\mathrm{~A}

Or

P_{\text{ave }}=\frac{I_{\max}V_{\max}}{2}

P_{\text{ave }}=\frac{I_{\max}V_{ms}\sqrt2}{2}

5236,36=\frac{l_{\max}(240)\sqrt2}{2}

I_{\max}=30,86\mathrm{~A}

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