Combinations of components - class-X

If we add two non ideal batteries in series then
so in case (a) the equivalent e.m.f.  may be larger than either of two e.m.f., but in case (b) the equivalent e.m.f. may be smaller than either of two e.m.f.
In series total resistance is always greater than individual resistance, whether batteries are connected in any way 

When two or more cells are connected in series in a circuit, the total potential difference is the sum of their potential differences. Hence, the cells are connected in series in order to produce a larger total emf.

In series, $I=\frac{nE}{nr}=\frac{E}{r}$ (i.e., I is 
independent of n)

In parallel, $I=\frac{E}{r/n}=\frac{nE}{r}, i.e,.\frac{nE}{r}, 
i.e.I\propto n$

After shorting by a wire having a constant resistance 
$I=E\left (R+\frac{r}{n}  \right )$

The total emf of the 20 cells connected in series are $1.8\times 20=36 V$. 

The back emf$= V-IR$ 
$= 12-2 = 10V$
$10%$ of this emf is $1V$
Hence the current is $IR = 1\times1= 1A$
Hence the current drawn by the loaded motor is $1+2=3A$

For identical cells, the current in series is E/(R + nr) and in parallel is E/(R + r/n). Setting these equal for R=4 gives r=4.

$ \Rightarrow  (100-12) = I[3+R]$

$ \dfrac{88}{3+R} = 8 $

$ \Rightarrow  R = 8\Omega$

For cells in series, current is E / (R + nbr), and in parallel, current is E / (R + r/n). Equating the currents under the condition that external resistance is the same gives R = r. Thus, the external resistance must equal the internal resistance of a single cell.

Using the potentiometer principle, the potential drop across 26 cm of the wire is proportional to the EMF of the cell. Calculation: (2.2 * 26/50) / (1 + 0.1) * (1) = 1.22V.

When two non-ideal batteries are connected in series, their equivalent emf is
$E _{eq} = E _1 + E _2$
Hence, the equivalent emf is larger than either of the two emfs.
Also, their equivalent internal resistance is
$r _{eq} = r _1 + r _2$
Hence, the equivalent resistance is larger than either of the two internal resistances.
Therefore, statement A is correct but B is wrong.

Let emf of each cell is $E$.
As they are connected in series so current in circuit is $I=\dfrac{E+E}{r _1+r _2+R}=\dfrac{2E}{r _1+r _2+R}$ 

(A) Option in series 

Number of cells $=n$

In series, current, $i _{1} = \dfrac {2E}{2 + 2r}$
In parallel, current, $i _{2} = \dfrac {E}{2 + \dfrac {r}{2}} = \dfrac {2E}{4 + r}$
According to the question
Since, $i _{1} - i _{2} \Rightarrow \dfrac {2E}{4 + r} = \dfrac {E}{2 + 2r}$
$\Rightarrow r = 2\Omega$.

Total resistance in the circuit is, $R _T=36\times0.5+12=30\Omega$

It is a common fact that when current flows the circuit, connected with a magnetic compass, the magnetic needle in the compass will show deflection due to the magnetic effect of the current. Magnetic lines of forces are created around the coil of wire and this caused the needle to deflect according to the direction of the current.
In this case, when the number of cells in the circuit is increased, the current through the circuit is also increased. This is because, the increase in cells indicates more current in the circuit. 
Hence, as the current in the circuit is increased, the deflection in the magnetic compass also increase further.

The cell has constant emf$,$

Resistance is $R _1$ and $R _2$

$P = V _2 / R _1 + r$

$P = V _2 / R _2 - r$

$( R _1- R _2 ) / 2$

The internal resistance of the cell is $( R _1 - R _2 ) / 2$

Hence,

option $(A)$ is correct answer.