Question 8.3.1: Additional Answer Options

2022 - May/June - Paper 1 - Question 8

Or

Option 2

V_{\text{int }}=Ir

1,2=I(0,5)

\mathrm{I}=2,4\mathrm{~A}

\mathrm{R}_{\mathrm{ext}}=\frac{\mathrm{V}}{\mathrm{I}}

=\frac{8,8}{2,4}

=3,67\Omega\quad(3,667)

\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2} I_{R}=2,4-0,73

\frac{1}{3,67}=\frac{1}{R}+\frac{1}{12}

R=5,29\Omega(5,28)

Or

Option 3

V_{\text{int }}=10-8,8

=1,2\mathrm{~V}

V_{\text{int }}=Ir

1,2=I(0,5)

L=2,4\mathrm{~A}

I_{\text{series branch }}=\frac{V}{R}

=\frac{8,8}{8+4}

=0,73\mathrm{~A}(0,733)

I_{R}=2,4-0,73

=1,67\mathrm{~A}(1,667)

R=\frac{V}{I_{R}}

=\frac{8,8}{1,67}

=5,27\Omega(5,28)

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Physical Sciences
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