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Question 9.6: Additional Answer Options

2021 - May/June - Paper 1 - Question 9

9.1

Slip rings

Tips:

Component X shows slip rings, which is confirmed by the fact that the information states that the diagram is an AC generator.
A split ring is a single ring and slip rings have two rings.

1 mark

9.2

Allows the slip rings to rotate while maintaining contact with the external circuit.

Or

Transfer/conduct current to the external circuit.

Or

The connection between the external circuit and coil/slip rings/internal circuit.

Tip: component Y shows carbon brushes which are made from graphite. These brushes allow the slip rings to rotate while maintaining electrical contact with the external circuit, alternatively, they transfer current to the external circuit.

1 mark

9.3

According to the principle of electromagnetic induction an emf/current is induced as a result of the change in the magnetic flux linkage with the coil.

Or

When the coil rotates there is a change of magnetic flux linked/associated with the coil. According to the principle of electromagnetic induction, an emf/current is induced in the coil.

Or

There is relative motion between the conductor and the magnetic field.

2 marks

(1 x 2 marks)

9.4

P to Q

Tips:

If the coil rotates clockwise, there is an upward force acting on segment PQ of the coil.
Using Flemings's right-hand generator rule, and remembering that the magnetic field points towards the south, you can determine that the induced current will flow from P to Q.

2 marks

(1 x 2 marks)

9.5

Option 1

$T=\frac{1}{f}=\frac{1}{50}$

$=0,02\mathrm{~s}$

$\mathrm{t}=(1,5)(0,02)$

$=0,03\mathrm{~s}$

Tips:

You need to use the frequency of 50 Hz to calculate the period of 0,02s.
Time t is the time taken for 1,5 periods.

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

3 marks

(3 x 1 mark)

9.6

How to approach this question:

To calculate the energy, you first need the rms potential difference.

Option 1

Part 1

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

$=\frac{311}{\sqrt2}$

$=219.91\mathrm{~V}$

Part 2

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

Although this equation does not state Vrms, the average energy dissipated can be determined by using this equation.

$=\frac{(219,11)^2\left(60\right)}{100}$

$=29016,24\mathrm{~J}$

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

5 marks

(5 x 1 mark)

Or

Option 2

Part 1

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

$=\frac{311}{\sqrt2}$

$=219.91\mathrm{~V}$

Part 2

$I_{rms}=\frac{V_{rms}}{R}$

$=\frac{219,91}{100}$

$=2,2\mathrm{A}(2,199)$

Part 3

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

=(219,91)(2,2)(60)

$=29028,12\mathrm{~J}$

(29013,61-29028,12)

Or

Option 3

Part 1

$I_{rms}=\frac{V_{rms}}{R}$

$=\frac{219,91}{100}$

$=2,2\mathrm{A}(2,199)$

Part 2

$W=I^2R\Delta t$

$=\left(2,2^2\right)(100)(60)$

$=29040\mathrm{~J}$

(29013,61-29040)

Or

Option 4

Part 1

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

$=\frac{311}{\sqrt2}$

$=219.91\mathrm{~V}$

Part 2

$I_{rms}=\frac{V_{rms}}{R}$

$=\frac{219,91}{100}$

$=2,2\mathrm{A}(2,199)$

Part 3

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

=(219,11)(2,2)

$=483,605\mathrm{~W}$

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

$483,605=\frac{W}{60}$

$W=29016,30\mathrm{~J}$

Or

Option 5

Part 1

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

$=\frac{311}{\sqrt2}$

$=219.91\mathrm{~V}$

Part 2

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

$=\frac{(219,11)^2}{100}$

$=483,605\mathrm{~W}$

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

$483,605=\frac{W}{60}$

$W=29016,30\mathrm{~J}$

Or

Option 6

Part 1

$I_{rms}=\frac{V_{rms}}{R}$

$=\frac{219,91}{100}$

$=2,2\mathrm{A}(2,199)$

Part 2

$P_{\text{ave }}=I_{\text{ms }}^2R$

=(2,2)^2(100)

$=483,605\mathrm{~W}$

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

$483,605=\frac{W}{60}$

$W=29016,30\mathrm{~J}$

Related subjects & topics

Grade 12 Physical Sciences

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