Question 8.1.3: 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). checkcheck

Or

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

Or

The rate at which work is done.checkcheck

light bulb Tip: 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}check

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

\mathrm{R}=24\Omegacheck

light bulb 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.

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

3 marks

(3 x 1 mark)

8.1.3

 

light bulb 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}check

\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)check

=36\Omega

Or

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\}checkcheck

=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.

V=IRcheck

Or

\varepsilon=l(R+r)check

12=I(36+2)check

I=0,32A(0,316A)check

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

5 marks

(5 x 1 mark)

8.1.4

 

Option 1

V=IR

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

=0,316(26)check

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

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

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

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

 

Or

Option 2

V=IR

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}check

V=IR_{//}

=(0,316)(12)check

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

Or

Option 3

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

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

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

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

V=0.158(24)check

=3,79\mathrm{~V}check

light bulb Tip: 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.

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}check 

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

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

3 marks

(3 x 1 mark)

8.2.1

 

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

2 marks

(2 x 1 mark)

8.2.2

 

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

light bulb 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)

Or

Option 2

Part 1

\operatorname{Rext}=(\mathrm{Rs}+\mathrm{R}//)

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

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

\mathrm{R}//=12\Omega

\operatorname{Rext}=(24+12)

=36\Omega

Or

R_{\text{ext }}=R_{s}+\frac{R_1R_2}{R_1+R_2}

R_{ext}=\left\{24+\frac{(24)(24)}{48}\right\}

=36\Omega

Part 2

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

I^2(36+2)=\frac{(12)^2}{38}

I=0,32A(0,316)

Or

I^2\mathrm{R}I^2\mathrm{R}=\frac{V^2}{R}

\mathrm{I}^2\mathrm{R}^2=\mathrm{V}^2

V=IR

12=I(38)

I=0,316\mathrm{~A}

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