A secondary cell have source emf of 3 V. When this cell is connected to a load of 0.25 then load emf calculated is 2.10 V. The internal resistance of cell and power consumed are

Options

A. `0.107 Omega` and 7.56 W
B. `0.208 Omega` and 3.34 W
C. `1.234 Omega` and 6.28 W
D. None of the above

Answer

A

Explanation

Given, source emf, `E_(s)=3V`
Load emf, V = 210 V
Load resistance, R = 0.25
`therefore " "` Internal resistance, `r = E_(s)-(V)/(I)` or `r = R((E_(s)-V))/(IR)`
Internal resistance, `r=R((E_(s)-V))/(V)` or `r=(0.25(3-2.10))/(2.10)`
or `" " r = 0.107 Omega`
Power consumed, `P = I^(2)r=((V)/(R ))^(2)r`
`P =((2.10)/(0.25))^(2)xx0.107=7.56 W`