Question

Two batteries with e.m.f. $12V$ and $13V$ are connected in parallel across a load resistor of $10\Omega $. The internal resistances of the two batteries are $1\Omega $ and $2\Omega $ respectively. The voltage across the load lies between.
A. $11.4V$ and $11.5V$
B. $11.7V$ and $11.8V$
C. $11.6V$ and $11.7V$
D. $11.5V$ and $11.6V$

Answer 1

Hint: This is a case of parallel cell combination with non zero internal resistances connected across a load resistance. Calculate the total emf and total internal resistance. Then find the current and using suitable equations calculate the potential across the load resistance. The total potential across the load will be equal to the total potential of the battery after the potential drop by the internal resistance.


Formula Used:
Resultant resistance when two resistances are connected in parallel, ${R_{net}} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Net emf when two cells are connected in parallel, ${E_{net}} = \dfrac{{\dfrac{{{E_1}}}{{{R_1}}} + \dfrac{{{E_2}}}{{{R_2}}}}}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}}}$
Current $i = \dfrac{{{E_{net}}}}{{{R_{total}}}}$
Also net potential of the battery ${V_{net}} = {E_{net}} - i{R_{net}}$

Complete step by step answer:
Given ${R_1} = 1\Omega $, ${R_2} = 2\Omega $, ${R_{load}} = 10\Omega $ ,${E_1} = 12V$ and ${E_2} = 13V$
First we need to find out the net internal resistance.
$ \Rightarrow {R_{net}} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Now we put the values of the known quantities. After than we get,

$ \Rightarrow {R_{net}} = \dfrac{{1 \times 2}}{{1 + 2}}$
$ \Rightarrow {R_{net}} = \dfrac{2}{3}\Omega $
Next we find net emf
$ \Rightarrow {E_{net}} = \dfrac{{\dfrac{{{E_1}}}{{{R_1}}} + \dfrac{{{E_2}}}{{{R_2}}}}}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}}}$
Now we put the values of the known quantities. After than we get,
$ \Rightarrow {E_{net}} = \dfrac{{\dfrac{{12}}{1} + \dfrac{{13}}{2}}}{{\dfrac{1}{1} + \dfrac{1}{2}}}$
$ \Rightarrow {E_{net}} = \dfrac{{\dfrac{{37}}{2}}}{{\dfrac{3}{2}}}$
$ \Rightarrow {E_{net}} = \dfrac{{37}}{3}V$
Therefore we find the Total resistance in the circuit.
$ \Rightarrow {R_{total}} = {R_{net}} + {R_{load}}$
$ \Rightarrow {R_{total}} = 10 + \dfrac{2}{3}$
$ \Rightarrow {R_{total}} = \dfrac{{32}}{3}\Omega $
Thus the current in the circuit is
$ \Rightarrow i = \dfrac{{{E_{net}}}}{{{R_{total}}}}$
$ \Rightarrow i = \dfrac{{\dfrac{{37}}{3}}}{{\dfrac{{32}}{3}}}$
\[ \Rightarrow i = \dfrac{{37}}{{32}}A\]
We know that the total potential across the load will be equal to the total potential of the battery after the potential drop by the internal resistance.
$ \Rightarrow {V_{net}} = {E_{net}} - i{R_{net}}$
Finally we put all the derived values in the above equation.
$ \Rightarrow {V_{net}} = \dfrac{{37}}{3} - \dfrac{{37}}{{32}} \times \dfrac{2}{3}$
$ \Rightarrow {V_{net}} = 11.5625V$
Hence option (D) is correct.

Note: Do all calculations properly. Also keep in mind that the cells are connected in parallel. Use the formula correctly. Do not forget that the total potential across the load will be equal to the total potential of the battery after the potential drop by the internal resistance.