A typical 12V automobile battery has a resistance of 0.012 Ω. What is the terminal voltage of this battery when the starter draws a current of 100 A? calculate the R, Pε, `\P_R` and `\P_r`. Answer (10.8 V, 0.108 N, 1200 w, 120 w)  



Given:

Battery e.m.f =  12 V

Resistance = R' = 0.012 Ω 

Current = I = 100 A

To Find:

1. Terminal Voltage = Vₜ = ?

2. Resistance = R = ?

3. Power of the Battery = Pε = ?

4. Power at load at resistance R = `\P_R` = ?

5. Power consumed at resistance r = `\P_r` = ?


Solution:

1. Terminal Voltage = Vₜ = ?

By using formula

e.m.f =  Vₜ + I R'

or 

Vₜ  = e.m.f - I R'

by putting values

Vₜ  = 12 V - 100 A ｘ 0.012 Ω 

Vₜ  = 12 V + 1.2 V 

 Vₜ  = 10.8 V ---------------Ans. (1)

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2. Resistance = R = ?

R =`\frac {Vₜ }{I}`

putting values

R =`\frac {10.8 V }{100 A}`

R =0.108 Ω -------------------Ans. (2)

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3. Battery Power = Pε = ?

Pε = I² ( R + R' )

putting values

Pε = (100 A)² ( 0.108 Ω + 0.012 Ω )

Pε = 10000 A² ｘ 0.12 Ω 

Pε = 1200 w ------------------Ans.(3)

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4. Power at load at resistance R = `\P_R` = ?

`\P_R` = I² R

putting values

`\P_R` = (100 A)² ( 0.108 Ω )

`\P_R` = 10000 A² ｘ 0.108 Ω 

`\P_R` = 1080 w  ------------------Ans.(4)

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5. Power consumed at resistance r = `\P_r` = ?


`\P_r` = I² R

putting values

`\P_r`  = (100 A)² ( 0.012 Ω )

`\P_r` = 10000 A² ｘ 0.012 Ω 

`\P_r`  = 120 w __________________Ans. (5)





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