(a) To get an equivalent resistance of $13.5 \Omega$, the resistances should beconnected as shown in the figure given below :So,$\;\frac{1}{R_P}\;=\;\frac{1}{R_1}+\;\frac{1}{R_2}$$\;=\;\frac{1}{9}+\;\frac{1}{9}=\;\frac{1+1}{9}=\;\frac{2}{9}$$\;\frac{1}{R_P}\;=\;\frac{2}{9}$$R_P\;=\;\frac{9}{2}=4.5 \Omega$$R_S\;= R_3+4.5 \Omega$$\;=9 \Omega+4.5 \Omega$$\;=13.5 \Omega$Now,$R_S\;= R_3+4.5 \Omega$$\;=9 \Omega+4.5 \Omega$$\;=13.5 \Omega$ (b) To get an equivalent resistance of $6 \Omega$, the resistances should be connected as shown in the figure : $R_S\;= R_1+ R_2$$\;=9+9$$\;=18 \Omega$Now both the resistors are in parallel with each other so,$R_P\;=\;\frac{1}{18}+\;\frac{1}{9}$$\;=\;\frac{1+2}{18}=\;\frac{3}{18}$$\;=\;\frac{1}{6} \Omega$So, $R_P=6 \Omega$OR a() According to Joule's law of heating, the heat produced in a wire is directly proportional to(i) square of current $(I^2)$,(ii) resistance of wire (R),(iii) time ( $t$ ) for which current is passed.Thus, the heat produced in the wire by current $I$ in time ' $t$ ' isor$H \propto I^2 R t$$H = (K I)^2 R t$$\text{ But } K=1 \text{, }$$P = V I$ (b) We know that, $\Rightarrow$$I = \frac{P}{V}$$P_1 = 100 \mathrm{W}, V = 220 \text{ volt }$First lamp :$I_1 = \frac{P_1}{V} = \frac{100}{220} = 0.45 \mathrm{A}$$P_2 = 60 \mathrm{W}, V = 220$Second lamp : $I_2 = \frac{P_2}{V} = \frac{60}{220} = 0.27 \mathrm{A}$ volt$\;= I_1 + I_2$So, Total current$\;= 0.45 + 0.27$$\;= 0.72 \mathrm{A}$$\;=0.72 {A}$