While doing an experiment with potentiometer (figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one and $A$ of the wire, to the end $R$; (ii) the deflection increased, while the jockey was moved towards the end $D$.
(i) Which terminal positive or negative of the cell $E_1$ is connected at $X$ in case (i) and how is $E_1$, related to $E$ ?
(ii) Which terminal of the cell $E_1$ is connected at $X$ in case (1 in 1 )?
(i) The deflection in galvanometer is one sided and the deflection decreased, while moving from one end ' $A$ ' of the wire to the end ' $B$ ', thus imply that current in auxiliary circuit (lower circuit containing primary cell) decreases, while potential difference across $A$ and jockey increases.
This is possible only when positive terminal of the cell $E_1$, is connected at $X$ and $E_1>E$.
(ii) The deflection in galvanometer is one sided and the deflection increased, while moving from one end $A$ of the wire to the end $B$, this imply that current in auxiliary circuit (lower circuit containing primary cell) increases, while potential difference across $A$ and jockey increases.
This is possible only when negative terminal of the cell $E_1$, is connected at $X$.
A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$. Plot a graph showing the variation of potential differential across $R$, versus $R$.
The graphical relationship between voltage across R and the resistance R is given below
First a set of $n$ equal resistors of $R$ each are connected in series to a battery of emf $E$ and internal resistance $R, A$ current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ' $n$ ' ?
In series combination of resistors, current $I$ is given by $I=\frac{E}{R+n R^{\prime}}$
whereas in parallel combination current $10 I$ is given by $$\frac{E}{R+\frac{R}{n}}=10 I$$
Now, according to problem,
$$\frac{1+n}{1+\frac{1}{n}} \Rightarrow 10=\left(\frac{1+n}{n+1}\right) n \Rightarrow n=10$$
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 $V_B>V_A$.
$$\begin{aligned} \Rightarrow\quad V_B-4 V-(0.2) \times 8 & =V_A \\ \text{Therefore}, \quad V_B-V_A & =3.6 \mathrm{~V} \end{aligned}$$