Question

The net emf of the three batteries as shown in the below figure is:

A. $1V$
B. $2V$
C. $6V$
D. $4V$

Answer 1

Hint: In electrical circuits, when two batteries are connected in series with each other, the net emf will get added in case of positive pole of one battery connected with negative pole of another batter whereas in case when positive pole of one battery connected with positive pole of another battery, net emf gets subtracted.

Formula used:
When two batteries of emf say ${\epsilon }_1$ and ${\epsilon }_2$ and having internal resistances of $r_1(and)r_2$ are connected in parallel combination, their net emf can be calculated as ${\epsilon }_{net}={\displaystyle \frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}}$ so, we will use these formulas to find net emf of the given circuit.

Complete step by step answer:
Firstly let us find the net emf of two batteries having emf’s of $2V(and)6V$ having internal resistances as shown in diagram of $1\mathrm{\Omega }$ each, are connected in parallel combination, but both batteries are connected with their same poles facing each other hence, the parallel formula will became as
${\epsilon }_{net}={\displaystyle \frac{{\epsilon }_1{r}_2-{\epsilon }_2{r}_1}{r_1+r_2}}$
Here, ${\epsilon }_1=6V$
${\epsilon }_2=2V$ And $r_1=r_2=1\mathrm{\Omega }$ so we get,
$\Rightarrow {\epsilon }_{net}={\displaystyle \frac{6-2}{2}}$
$\Rightarrow {\epsilon }_{net}=2V$
With polarity of negative terminal facing to the negative terminal of battery having emf of $4V$. Now, the battery of emf $4V$ and this net emf of ${\epsilon }_{net}=2V$ both are connected in series to each other, so their net emf can be calculated as:
${\epsilon }_{total}=4-{\epsilon }_{net}$
$\Rightarrow {\epsilon }_{total}=4-2$
So, net emf of three batteries is ${\epsilon }_{total}=2V$

Hence, the correct option is B.

Note:It should be remembered that, while calculating net emf between two batteries, check the facing polarity of two batteries terminal if they have same polarity facing to each other, then emf gets subtracted and they are facing opposite polarity with each other, net emf will get added.