Question

A battery made of 5 cells each of 2 V and having internal resistance of 0.1 Ω, 0.2 Ω, 0.3 Ω, 0.4 Ω and 0.5 Ω is connected across 10 Ω resistance. Draw circuit diagram and calculate the current flowing through 10 Ω resistance.​

Answer 1

Answer

see the attachment !

Answer 2

Answer:  I = 0.869 A

Explanation:

Given,

That, 5 cells having internal resistance of 0.1 Ω , 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 and a resistance of 10 Ω are connected in the series.

Hence,

Total resistance of the circuit,

R = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 10

R = 1.5 + 10

R = 11.5 Ω

Given that 5  cells are connected in the series, Applying the second Kirchhoff's in the closed loop, Total voltage,

V = 2 + 2 + 2 + 2 + 2

V = 10 volts.

Using ohm's law,

Current in circuit,

I = V/R

I = 10/11.5

I = 0.869 A