Question

Six batteries of increasing emf and increasing internal resistance are as shown in figure.

Table 1	Table - 2
(a) Potential of point A	(p) Zero
(b) Potential of point B	(q) 2 V
(c) Potential of point C	(r) 4 V
(d) Potential of point D	(s) 6 V

A. a - p; b - p; c - p,d-p
B. a-p,b-s,c-q
C. a-q,b-q,c-s
D. a-r,b-r,c-p

Answer 1

Hint: In order to find the solution of above numerical we need to understand the diagram, according to the figure the six batteries are connected such that increases in internal resistance and emf. We need to use the formula of potential so we can easily solve this problem.

Complete step by step answer:
From the above diagram WKT the potential difference of ground is equal,
${V_{Ground}} = {V_{Ground}}$
The equivalent resistance is equal to the emf and I be the current flowing through the circuit
We calculate the current by using,
${R_{eq}} = {E_{eq}}$
$IV = {E_{eq}}$
$\
\implies I(21) = 21 \\
\implies I = 1\Omega \\
\ $
To calculate the potential difference through point A is
By using this formula, we apply this formula for each case
$emf = (R + r)I$
$\
  {V_A} - {V_G} = - 2 + 2 \times 1 - 3 + 3 \times 1 \\
\implies {V_A} - {V_G} = 0 \\
\implies {V_A} = 0 \\
\ $
To calculate the potential difference through point B is
$\
  {V_B} - {V_G} = - 3 + 3 \times 1 \\
\implies {V_B} - {V_G} = 0 \\
\implies {V_B} = 0 \\
\ $
To calculate the potential difference through point C is
$\
  {V_C} - {V_G} = 1 + 1 \times 1 - 2 + 2 \times 1 - 3 + 3 \times 1 \\
\implies {V_C} - {V_G} = 0 \\
\implies {V_C} = 0 \\
\ $
To calculate the potential difference through point D is
$\
  {V_D} - {V_G} = 5 - 1 \times 5 + 4 - 1 \times 4 \\
\implies {V_D} - {V_G} = 0 \\
\implies {V_D} = 0 \\
\ $

So, the correct answer is “Option A”.

Note:
The terminal potential difference across the cell is always less than emf of a cell.
Commonly the Internal resistance is the opposition offered by the flow of charges by the cells and batteries are responsible for generating the heat.