Question 8.3.2: Additional Answer Options

2021 - May/June - Paper 1 - Question 8



(Electrical) energy/work check

Unit charge check

light bulb Tip: the emf of the battery is the maximum energy/work provided by a battery per unit of charge passing through it.

2 marks

(2 x 1 mark)




light bulbHow to approach this question: 

Option 1

  • Calculate the total resistance of the top branch of the circuit.


=4+3 check


  • Next, calculate the total effective parallel resistance.

\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2} check

\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac17+\frac17 check

R_{p}=3,5\Omega check


  • While normally marks aren't awarded for simply showing the substitution of resistors in series, in this case, you should be awarded a mark for using 7Ω as the resistance for the top branch.
  • It is always a good idea to show all your work, no matter how basic it may seem. This can help you avoid losing marks in cases where partial credit is awarded for the process.

4 marks

(4 x 1 mark)




light bulb How to approach this question: 

  • You will need to do simultaneous equations to solve for emf and r since there are 2 unknown variables.
  • It is possible to calculate the current using the external potential difference of 2,8V and the external resistance of 7Ω.
  • Remember, when the switch is open, the current only flows through one branch of the parallel combination.

When the switch is open:

I=\frac{V}{R} check(1)

\mathrm{I}=\frac{2,8}{7} check (2)


  • It is possible to use either formula below to calculate the first of the two equations needed to determine the internal resistance and emf.

\varepsilon=I(R+r) check(4)

=0,4(7+r) check(5)


\varepsilon=V_{\text{ext }}+\mathrm{lr}check(4)


When the switch is closed:

  • The total resistance of the parallel combination is 3,5 Ω as calculated in question 8.2 and the potential difference is 2,63V.

I=\frac{V}{R} check(1)

\mathrm{I}=\frac{2,63}{3,5} check(3)


\varepsilon=I(R+r) check(4)



\varepsilon=V_{\text{ext }}+\mathrm{lr}check(4)

\varepsilon=2,63+(0,75)r check(6)

  • Equating the two equations allows you to solve for the internal resistance.

2,8+(0,4)r=2,63+(0,75)r check(7)

r=0,49\Omega check(8)


0,4(7+r)=0,75(3,5+r) check(7)

r=0,49\Omega check(8)

8 marks

(8 x 1 mark)

Marking criteria:

Calculation of current when the switch is open and when closed:

  • (1) Suitable formula for open or closed switches check
  • (2) Correct substitution when the switch is open check
  • (3) Correct substitution when the switch is close check

Substitution into formula ε = I(R + r) or ε = Vext + Ir:

  • (4) Formula check
  • (5) Substitution in the formula for an open switch check
  • (6) Substitution in the formula for a closed switch check

Calculating r:

  • (7) Equating the equations check
  • (8) Final answer: 0,49 Ω check




Option 1

light bulb Tip: the marks are awarded for the correct substitution of the value of the internal resistance into any formula used in question 8.3.1

\varepsilon=\mathrm{V}_{\text{ext }}+\mathrm{Ir}




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

2 marks

(2 x 1 mark)



Option 2

\varepsilon=V_{\text{ext }}+\mathrm{Ir}





Option 3





Option 4




Related subjects & topics
Explore similar posts in our community