Question

A battery having an emf of 9.63 V delivers 118 mA when connected to a 60.0 Ω load. Determine the internal resistance of the battery.

Answer 1

21.6 ohm

Step-by-step explanation:

EMF of the battery, E = 9.63 V

Current, i = 118 mA = 0.118 A

Resistance, R = 60 ohm

Let the internal resistance of the cell is r.

`i = (E)/(R + r)`

R + r = 9.63 / 0.118

60 + r = 81.6

r = 21.6 ohm

Answer 2

21.6 ohm

Step-by-step explanation:

We are given that

EMF=E=9.63 V

Current=I=118 mA=
`118* 10^(-3) A`

`1 mA=10^(-3) A`

Resistance=
`R=60\Omega`

We have to find the internal resistance of the battery.

We know that

`V=E-Ir`

We know that V=IR

`IR=E-Ir`

`IR+Ir=E`

`I(R+r)=E`

`R+r=(E)/(I)`

Substitute the values

`60+r=(9.63)/(118* 10^(-3))`

`60+r=81.6`

`r=81.6-60`

`r=21.6\Omega`

Hence, the internal resistance of the battery=21.6 ohm