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

Question

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

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
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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
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`