Question

Four cells of emf 1.5v and internal resistance 0.5ohm connected in series but one cell is wrongly connected.the net voltage and net internal resistance between A and B is?

Answer 1

Refer to answer and mark BRAINLIEST

Answer 2

net voltage between A and B will be 3V and net internal resistance between A and B is  2 Ω

given

Four cells of emf 1.5v

internal resistance 0.5ohm

Four cells connected in series

one cell is wrongly connected

to find

net voltage and net internal resistance between A and B

solution

• it is given that one cell is wrongly connected , it means that positive and negative terminals of the battery are interchanged and are different from that of the other three.

• since the cells are connected in series . therefore the total EMF of all the connected cells is

E = e1 + e2 + e3 + e4

E = 1.5 + 1.5 + 1.5 - 1.5

E = 3V

• the resistance of all the cells will be calculated by observing the fact that all the cells are connected in series and their internal resistances are also connected in series .
• therefore the total internal resistance of the cells is given by:

R = r1 + r2 + r3 + r4

R = 0.5 + 0.5 + 0.5 + 0.5

R = 2 Ω

• so the net voltage between A and B will be 3V and the net internal resistance between A and B will be 2Ω.

net voltage between A and B will be 3V and net internal resistance between A and B is  2 Ω

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