Additional Answer Options: 8.2.1

2018 - November - Paper 1 - Question 8

8.1

The battery supplies 12 J per coulomb/12 J per unit charge. checkcheck

light bulbTips: 

  • EMF is the amount of energy (12J) per coulomb of charge that passes through the battery.
  • Your answer must explain the meaning of an emf of 12V, not simply the meaning of an emf.

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

2 marks

(1 x 2 marks)

8.2.1

Option 1

\mathrm{V}_{\text{lost }}=\mathrm{Ir}check

=(2)(0,5)

=1\mathrm{~V}

V_{\text{ext }}=Emf/emk-V_{\text{lost}}

=(12-1)check

=11\mathrm{~V}check

light bulb Tips: 

  • The lost volts are equal to the product of the internal resistance and the current.
  • The voltmeter reads the external potential difference, which is the difference between the EMF and the lost volts.

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

3 marks

(3 x 1 mark)

8.2.2

Option 1

R=\frac{V}{I}

=\frac{11}{2}check

=5,5\Omegacheck

light bulb Tip: the voltage across the resistor measures 11V and the reading on the ammeter (which measures the total current) is 2A.

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

2 marks

(2 x 1 mark)

8.3 

Decreases check

  • Total resistance decreasescheck
  • Current increasescheck
  • "Lost volts" increases, (emf the same)check

Or

Decreases check 

  • Total resistance decreasescheck 
  • Current increasescheck
  • ε = Vext + Ir Ir increases. ε is constant. ∴Vext decreasescheck

light bulb Tips: 

  • When the switch S is closed, it will add a resistor in parallel which decreases the total resistance of the circuit.
  • When the total resistance of the circuit decreases, the current increases; this increases the lost volts and decreases the voltmeter reading.

4 marks

(4 x 1 mark)


Or

Option 2

\varepsilon=I(R+r)

12=V_{\text{ext}}+(2)(0,5)

\mathrm{V}_{\text{ext}}=11\mathrm{V}


Or

Option 3

\varepsilon=\mathrm{l}(\mathrm{R}+\mathrm{r})

12=2(R+0,5)

\mathrm{R}=5,5\Omega

V=IR

=2(5,5)

=11\mathrm{~V}

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