Let there be $n$ resistors $R_1 \ldots \ldots . R_n$ with $R_{\max }=\max \left(R_1 \ldots \ldots \ldots R_n\right)$ and $R_{\min }=\min \left\{R_{1 . . .} \quad R_n\right\}$. Show that when they are connected in parallel, the resultant resistance $R_p=R_{\min }$ and when they are connected in series, the resultant resistance $R_s>R_{\max }$. Interpret the result physically.
When all resistances are connected in parallel, the resultant resistance $R_p$ is given by
$$\frac{1}{R_p}=\frac{1}{R_1}+\ldots \ldots . .+\frac{1}{R_n}$$
On multiplying both sides by $R_{\min }$ we have
$$\frac{R_{\min }}{R_p}=\frac{R_{\min }}{R_1}+\frac{R_{\min }}{R_2}+\ldots .+\frac{R_{\min }}{R_n}$$
Here, in RHS, there exist one term $\frac{R_{\min }}{R_{\min }}=1$ and other terms are positive, so we have
$$\frac{R_{\min }}{R_p}=\frac{R_{\min }}{R_1}+\frac{R_{\min }}{R_2}+\ldots .+\frac{R_{\min }}{R_n}>1$$
This shows that the resultant resistance $R_p
Thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors. Now, in series combination, the equivalent resistant is given by
$$R_s=R_1+\ldots \ldots+R_n$$
Here, in RHS, there exist one term having resistance $R_{\text {max }}$.
So, we have
$$\begin{aligned} \text{or}\quad & R_s=R_1+\ldots+R_{\max }+\ldots+\ldots+R_n \\ & R_s=R_1+\ldots+R_{\max \ldots}+R_n=R_{\max }+\supset\left(R_1+\supset+\right) R_n \\ \text{or}\quad & R_s \geq R_{\max } \\ & R_s=R_{\max }\left(R_1+\ldots+R_n\right) \end{aligned}$$
Thus, in series combination, the equivalent resistance of resistors is greater than the maximum resistance available in combination of resistors. Physical interpretation
In Fig. (b), $R_{\text {min }}$ provides an equivalent route as in Fig. (a) for current. But in addition there are $(n-1)$ routes by the remaining $(\mathrm{n}-1)$ resistors. Current in Fig. (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) $
In Fig. (d), $R_{\max }$ provides an equivalent route as in Fig. (c) for current. Current in Fig. (d) $<$ current in Fig. (c). Effective resistance in Fig. (d) $>R_{\max }$. Second circuit evidently affords a greater resistance.
The circuit in figure shows two cells connected in opposition to each other. Cell $E_1$ is of emf 6 V and internal resistance $2 \Omega$ the cell $E_2$ is of emf 4 V and internal resistance $8 \Omega$. Find the potential difference between the points $A$ and $B$.
Applying Ohm's law.
Effective resistance $=2 \Omega+8 \Omega=10 \Omega$ and effective emf of two cells $=6-4=2 \mathrm{~V}$, so the electric current is given by
$$I=\frac{6-4}{2+8}=0.2 \mathrm{~A}$$
along anti-clockwise direction, since $E_1>E_2$.
The direction of flow of current is always from high potential to low potential. Therefore $V_B>V_A$.
$$\begin{aligned} \Rightarrow\quad V_B-4 V-(0.2) \times 8 & =V_A \\ \text{Therefore}, \quad V_B-V_A & =3.6 \mathrm{~V} \end{aligned}$$
Two cells of same emf $E$ but internal resistance $r_1$ and $r_2$ are connected in series to an external resistor $R$ (figure). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero?
Applying Ohm's law,
Effective resistance $=R+r_1+r_2$ and effective emf of two cells $=E+E=2 E$, so the electric current is given by
$$I=\frac{E+E}{R+r_1+r_2}$$
The potential difference across the terminals of the first cell and putting it equal to zero.
$$\begin{aligned} & V=E-I_1=E-\frac{2 E}{r_1+r_2+R} r_1=0 \\ \text{or}\quad & E=\frac{2 E r_1}{r_1+r_2+R} \Rightarrow 1=\frac{2 r_1}{r_1+r_2+R} \\ & r_1+r_2+R=2 r_1 \Rightarrow R=r_1-r_2 \end{aligned}$$
This is the required relation.
Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 mm . Conductor $B$ is a hollow tube of outer diameter 2 mm and inner diameter 1 mm . Find the ratio of resistance $R_A$ to $R_B$.
The resistance of first conductor
$$R_A=\frac{\rho l}{\pi\left(10^{-3} \times 0.5\right)^2}$$
The resistance of second conductor,
$$R_B=\frac{\rho l}{\pi\left[\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2\right]}$$
Now, the ratio of two resistors is given by
$$\frac{R_A}{R_B}=\frac{\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2}{\left(0.5 \times 10^{-3}\right)^2}=3: 1$$
Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Examples 3,7 in the NCERT Text Book for Class XII.
Let the effective internal resistance of the battery is $R_{\text {eff }}$, the effective external resistance $R$ and the effective voltage of the battery is $V_{\text {eff }}$.
Applying Ohm's law,
Then current through $R$ is given by
$$I=\frac{V_{\text {eff }}}{R_{\text {eff }}+R}$$
If all the resistances and the effective voltage are increased $n$-times, then we have
$$\begin{aligned} V_{\text {eff }}^{\text {new }} & =n V_{\text {eff }}, R_{\text {eff }}^{\text {new }}=n R_{\text {eff }} \\ \text{and}\quad R^{\text {new }} & =n R \end{aligned}$$
Then, the new current is given by
$$I^{\prime}=\frac{n V_{\text {eff }}}{n R_{\text {eff }}+n R}=\frac{n\left(V_{\text {eff }}\right)}{n\left(R_{\text {eff }}+R\right)}=\frac{\left(V_{\text {eff }}\right)}{\left(R_{\text {eff }}+R\right)}=I$$
Thus, current remains the same.