(a) Let three resistors $R_1, R_2$ and $R_3$ are connected in series which are also connected with a battery, an ammeter and a key as shown in figure. When key is closed, the current starts flowing through the circuit. Take the reading of ammeter. Now change the position of ammeter to anywhere in between the resistors and take its reading. We will observe that in both the cases reading of ammeter will be same showing same current flows through every part of the circuit above. (b) Given,$\; R_1=5\ \Omega,\ R_2=10\ \Omega,\ R_3=15\ \Omega,\ V=30\ \mathrm{V}$Total resistance,$R= R_1+ R_2+ R_3$$[ \because 5\ \Omega,\ 10\ \Omega$and $15\ \Omega$are connected in series]$\; =5+10+15$$=30\ \Omega$Potential difference, $V=30\ \mathrm{V}$Current in the circuit, $I=$ ?From Ohm's law.$I=\frac{V}{R}=\frac{30\ \mathrm{V}}{30\ I}=1\ \mathrm{A}$$\therefore$ Current flowing in the circuit $=1\ \mathrm{A}$Potential difference across $15\ \Omega$ resistors $=\mathbb{R}_3$$\;=1\ \mathrm{A} \times 15\ \Omega$$\;=15\ \mathrm{V}$OR (a) Let $R_1, R_2$ and $R_3$ are three resistance connected in parallel to one another and $R$ is the equivalent resistance of the circuit. A battery of V volts has been applied across the ends of this combination. When the switch of the key is closed, current I flows in the circuit such that,From Ohm's law, $\;I=\frac{V}{R}$...(i)$\;I_1=\frac{V}{R_1}$...(ii)$\;I_2=\frac{V}{R_2}$...(iii)$\;I_3=\frac{V}{R_3}$...(iv)$\;I=I_1+I_2+I_3$...(v) Putting the values of $I,\ I_1,\ I_2$ and $I_3$ in equation (v), $\; \frac{V}{R}= \; \frac{V}{R_1}+ \; \frac{V}{R_2}+ \; \frac{V}{R_3}$ $V \left( \; \frac{1}{R} \right)=V \left( \; \frac{1}{R_1}+ \; \frac{1}{R_2}+ \; \frac{1}{R_3} \right)$ $\; \frac{1}{R}= \; \frac{1}{R_1}+ \; \frac{1}{R_2}+ \; \frac{1}{R_3}$ (b) Let $R_P$ is the equivalent resistance of resistors connected in parallel. $\therefore$ Equivalent resistance of resistors in parallel:$\; \frac{1}{R_P} \; = \; \frac{1}{20}+ \; \frac{1}{20}$$\; \frac{1}{R_P} \; = \; \frac{1+1}{20}= \; \frac{2}{20}= \; \frac{1}{10}$$R_P \; =10\ \Omega .$Now, equivalent circuit becomes.$\because \; \; 10\ \Omega$ and $10\ \Omega$ are connected in series.$\therefore \; \;$ Equivalent resistance of the circuit$\; =10\ \Omega+10\ \Omega$$\; =20\ \Omega$