(a) It is observed that total current I is equal to the sum of separate currents.$I = I_1 + I_2 + I_3$....(i)Let $R_p$ be the equivalent resistance of the parallel combination of resistors.So, by applying ohm's law $\; I = \frac{V}{R_p}$$\; I_1 = \frac{V}{R_1}, I_2 = \frac{V}{R_2} \;\text{ and }\; I_3 = \frac{V}{R_3}$ So, now from equation (i), we have$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$ and$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ Hence, if $n$ resistors are connected in parallel, then the equivalent resistance of the circuit is given by -$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$ (b) Given, Two resistors of $12\ \Omega$ connected in parallel.$\because \frac{1}{R_{\text{eq}}} = 6\ \mathrm{V}$$\therefore \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{12} + \frac{1}{12}$$\frac{1}{R_{\text{eq}}} = \frac{2}{12}$$R_{\text{eq}} = \frac{12}{2} = 6\ \Omega$ According to ohm's law V=IR$6\; = I \times 6$$\frac{6}{6} = I$$I\; = 1 \;\text{ ampere. }\;$OR (a) Given, $R_1 = 20\ \Omega, R_2 = 4\ \Omega$Since, in Series$R = R_1 + R_2$$\therefore$ Total resistance of circuit :$R\; = 20 + 4$$\; = 24\ \Omega$(b) Current through circuit :$V = 6\ \mathrm{V}, R = 24\ \Omega$According to ohm's law$V = IR$So,$I = \frac{V}{R}$ $I = \frac{6}{24} = \frac{1}{4} = 0.25$ ampere (c) (i) Potential difference across conduction: $I\; = \frac{1}{4}, R = 20\ \Omega$$V_1\; = I_1$$V_1\; = \frac{1}{4} \times 20 = 5\ \mathrm{V}$ (ii) Potential difference across lamp $V_2 = I R_2$$\; = \frac{1}{4} \times 4$$V_2 = 1\ \mathrm{V}$ (d) Power of lamp :$P = I^2 R$$\; = \left(\frac{1}{4}\right)^2 \times 20$$\; = \frac{1}{4} \times \frac{1}{4} \times 20 = \frac{5}{4}\ \mathrm{watt}$or$P = 1.25 \;\text{ watt }\;$