Solveeit Logo

Question

Question: The net emf of the three batteries as shown in the below figure is: ![](https://www.vedantu.com/qu...

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

A. 1V1V
B. 2V2V
C. 6V6V
D. 4V4V

Explanation

Solution

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 ε1{\varepsilon _1} and ε2{\varepsilon _2} and having internal resistances of r1(and)r2{r_1}(and){r_2} are connected in parallel combination, their net emf can be calculated as εnet=ε1r2+ε2r1r1+r2{\varepsilon _{net}} = \dfrac{{{\varepsilon _1}{r_2} + {\varepsilon _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)6V2V(and)6V having internal resistances as shown in diagram of 1Ω1\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
εnet=ε1r2ε2r1r1+r2{\varepsilon _{net}} = \dfrac{{{\varepsilon _1}{r_2} - {\varepsilon _2}{r_1}}}{{{r_1} + {r_2}}}
Here, ε1=6V{\varepsilon _1} = 6V
ε2=2V{\varepsilon _2} = 2V And r1=r2=1Ω{r_1} = {r_2} = 1\Omega so we get,
εnet=622\Rightarrow {\varepsilon _{net}} = \dfrac{{6 - 2}}{2}
εnet=2V\Rightarrow {\varepsilon _{net}} = 2V
With polarity of negative terminal facing to the negative terminal of battery having emf of 4V4V. Now, the battery of emf 4V4V and this net emf of εnet=2V{\varepsilon _{net}} = 2V both are connected in series to each other, so their net emf can be calculated as:
εtotal=4εnet{\varepsilon _{total}} = 4 - {\varepsilon _{net}}
εtotal=42\Rightarrow {\varepsilon _{total}} = 4 - 2
So, net emf of three batteries is εtotal=2V{\varepsilon _{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.