Question 8.1.5: Additional Answer Options

2019 - May/June - Paper 1 - Question 8

8.1.1

The rate at which (electrical) energy is converted (to other forms) (in a circuit).

The rate at which energy is used / Energy used per second.

The rate at which work is done.

The amount of work done or energy transferred per unit of time has to be clear.

2 marks

(1 x 2 marks)

8.1.2

Option 1:

$P=\frac{V^2}{R}$

$6=\frac{(12)^2}{R}$

$\mathrm{R}=24\Omega$

Tips:

The most direct method to calculate the resistance is to use the power of 6W and the potential difference of 12V as shown above.
Other methods can be used to get to the correct answer.

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3 marks

(3 x 1 mark)

8.1.3

How to approach this question:

The first step is to calculate the effective resistance of the parallel resistors using the values calculated in question 8.1.2.

Option 1:

Part 1:

$\frac{1}{R_{//}}=\frac{1}{R_1}+\frac{1}{R_2}$

$=\frac{1}{24}+\frac{1}{24}$

$\mathrm{R}_{//}=12\Omega$

The total resistance is the sum of the parallel resistors B and C as well as the resistor A.

$R_{ext}=\left(R_{s}+R_{//}\right)$

$R_{ext}=(24+12)$

$=36\Omega$

$R_{\text{total }}=\left(R_{s}+\frac{R_1R_2}{R_1+R_2}\right)$

$R_{\text{total }}=\left\{24+\frac{(24)(24)}{48}\right\}$

$=36\Omega$

Part 2:

The current can be calculated using the emf of 12V, the external resistance of 36 Ω and the internal resistance of 2Ω.

$\varepsilon=l(R+r)$

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5 marks

(5 x 1 mark)

8.1.4

How to approach this question:

The potential difference across the parallel components can be determined by subtracting the product of the current and the sum of the resistance of A as well as the internal resistance from the emf.

Option 1:

$\mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{A}}+\mathrm{r}\right)$

$=8,216\vee(8,32\vee)$

$V_{//}=(12-8,216)$

$=3,784\mathrm{~V}(3,68\mathrm{~V})$

$\therefore\mathrm{V}_{\mathrm{C}}=3,78\mathrm{~V}(3,68\mathrm{~V})$

Option 2:

For the parallel portion:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\text{ OR }R=\frac{R_1R_2}{\left(R_1+R_2\right)}$

$R=\frac{(24)(24)}{48}$

$=12\Omega$

$V_{//}=V_{C}$

$V=IR_{//}$

$=3.79\mathrm{~V}(3,84\mathrm{~V})$

Option 3:

$I_{A}=I_{B}+I_{C}$

$=2\mathrm{I}_{\mathrm{B}}$

$0,316=2\mathrm{I}_{\mathrm{B}}$

$\mathrm{I}_{\mathrm{B}}=0,158\mathrm{~A}$

$=3,79\mathrm{~V}$

3 marks

(3 x 1 mark)

8.1.5

Option 1:

The power rating (output voltage) of the bulb is 6 W, 12 V.

$\mathrm{P}=\frac{V^2}{R}$

For a given resistance, power is directly proportional to V².
Since the potential difference across light bulb C calculated in question 8.1.5 is less than the operating voltage of 12V, the output/power will be less.

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3 marks

(3 x 1 mark)

8.2.1

The total current passes through resistor A. For the parallel portion, the current branches, therefore, only a portion of the total current passes through resistor C.

2 marks

(2 x 1 mark)

8.2.2

The current in B is equal to the current in A. The circuit becomes a series circuit.

Tips:

The current is the same at all points in a series circuit.
When resistor C is removed, resistors A, B and D are all connected in series.

2 marks

(2 x 1 mark)

Option 2

$\mathrm{P}=\frac{V^2}{R}$

The potential difference across light bulb C is less than the operating voltage. Thus for the same resistance, brightness decreases.

Option 3

P=1^2R

In the circuit, the total current in light bulb C is less than the optimum current required (0,5 A). Thus for the same resistance, the power will be less and, therefore, the brightness will decrease.

Option 4

P=1^2R

For a given resistance, power is directly proportional to I2. Since the current decreases, brightness decreases.

Option 5

P=IV

Power is directly proportional/equal to the product of V and I. Since the current decreases, brightness decreases.

The voltage across light bulb C, as well as the current in the bulb, are all less than the optimum values, therefore, the power and brightness are less.