(A) Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of 9 Ω. Also justify your answer.

When two \(\displaystyle 6 \Omega\) resistances are connected in parallel and the third resistance of \(\displaystyle 6 \Omega\) is connected in series combinations to this, then equivalent resistance will be \(\displaystyle 9 \Omega /\)

Equivalent resistance \(\displaystyle =R_{1}+R_{2}=1 \Omega+2 \Omega=3 \Omega\)

\[\begin{aligned} & I=\frac{\mathrm{V}^{7}}{{ }_{6 \mathrm{~V}}} \\ & =\frac{6 \mathrm{~V}}{1 \Omega+2 \Omega}=\frac{6 \mathrm{~V}}{3 \Omega}=2 \mathrm{~A} \end{aligned} \]

Electric power, \(\displaystyle \mathrm{P}=\mathrm{I}^{2} \mathrm{R}\)

\[=(2 \mathrm{~A})^{2} \times 2 \Omega=4 \times 2 \mathrm{~W}=8 \mathrm{~W} \]