Question 9.2.2: Additional Answer Options

2019 - November - Paper 1 - Question 9

9.1.1

 

DC check

light bulb Tip: the generator has a split ring, therefore, it is a direct current generator.

1 mark

 

9.1.2

 

Emf is induced as a result of a change in magnetic flux (linked) with the coil.checkcheck

light bulb Tip: the principle of electromagnetic induction is that an emf is induced when there is a change in the magnetic flux linkage.

2 marks

(1 x 2 marks)

9.1.3

checkcheck 

 

Or

checkcheck 

 

Or

checkcheck 

 

Or

checkcheck 

 

light bulb Tips:

  • A DC generator will produce a current that only flows in one direction and a potential difference that is always positive or always negative.
  • One complete cycle results in two maxima or two half cycles. It can start at a voltage of zero or the maxima.

2 marks

(1 x 2 marks)

Marking criteria:

  • Correct shapecheck
  • One complete cyclecheck

 

9.2.1

The AC potential difference/voltage dissipates the same amount of energy as the DC potential difference. checkcheck

Or

(The rms value of AC is) the DC potential difference/voltage which dissipates the same amount of energy as the AC potential difference/voltage. checkcheck

light bulb Tip: the rms voltage is the potential difference that, in an AC circuit dissipates the same amount of energy as a DC circuit with the same potential difference.

2 marks

(1 x 2 marks)

9.2.2

Option 1

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

500=\frac{V^2}{200}(10)check

\mathrm{V}=\mathrm{V}_{\mathrm{rms}}=100\mathrm{~V}

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

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

V_{\text{max }}=141,42\mathrm{~V}check

light bulb Tips:

  • The energy dissipated by the resistor is 500J, and the resistor has a resistance of 200Ω.
  • It is possible to work out the Vrms and to use this to calculate the Vmax.

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

5 marks

(5 x 1 mark)

Or

Option 2

W=I^2R\Delta t

500=I^2(200)(10)

I=I_{\mathrm{rms}}=0,5\mathrm{~A}

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

\frac{500}{10}=V_{\mathrm{rms}}(0,5)

\mathrm{V}_{\mathrm{rm}}=100\mathrm{~V}

V_{\mathrm{rms}}=\frac{V_{\max}}{\sqrt2}

100=\frac{V_{\max}}{\sqrt2}

V_{\max}=141,42\mathrm{~V}

Or

Option 3

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

\frac{500}{10}=\mathrm{I}_{\mathrm{rms}}^2

\text{ \lparen200\rparen }

\mathrm{I}_{\mathrm{rms}}=0,5\mathrm{~A}

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

\frac{500}{10}=V_{\text{rms }}\left(0,5\right)

\mathrm{V}_{\mathrm{rms}}=100\mathrm{~V}

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

100=\frac{V_{\max}}{\sqrt2}

V_{\max}=141,42\mathrm{~V}

Or

Option 4

\mathrm{R}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{rms}}}

200=\frac{V_{\text{rms }}}{0,5}

\mathrm{V}_{\mathrm{rms}}=100\mathrm{~V}

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

100=\frac{V_{\max}}{\sqrt2}

V_{\max}=141,42\mathrm{~V}

Or

Option 5

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

\frac{500}{10}=\frac{V_{\mathrm{rms}}^2}{200}

\mathrm{V}_{\mathrm{rms}}=100\mathrm{~V}

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

100=\frac{V_{\max}}{\sqrt2}

V_{\max}=141,42\mathrm{~V}

Or

Option 6

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

\frac{500}{10}=\frac{P_{\max}}{2}

P_{\max}=100\mathrm{~W}

P_{\max}=\frac{V_{\max}^2}{R}

100=\frac{V_{\max}^2}{200}

V_{\max}=141,42\mathrm{~V}

Or

Option 7

P=\frac{W}{\Delta t}=\frac{500}{10}

=50\mathrm{~W}

P_{\mathrm{ave}}=\frac{\mathrm{V}_{\mathrm{rms}}^2}{\mathrm{R}}

50=\frac{V_{\mathrm{rms}}^2}{200}

\mathrm{V}_{\mathrm{rms}}=100\mathrm{~V}

V_{\mathrm{rms}}=\frac{V_{\max}}{\sqrt2}

100=\frac{V_{\max}}{\sqrt2}

V_{\max}=141,42\mathrm{~V}

Or

Option 8

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

\frac{500}{10}=\mathrm{I}_{\mathrm{rms}}^2(200)

I_{\mathrm{rms}}=0,5\mathrm{~A}

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

0,5=\frac{I_{\max}}{\sqrt2}

I_{\max}=0,71\mathrm{~A}

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

=(0,71)(200)

=142\mathrm{~V}

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