An ammeter with resistance 1.42 Ω is connected momentarily across a battery, and the meter reads 9.73 A . When the measurement is repeated with a 2.11-Ω meter, the reading is 7.36 A .

Find

(a) the battery voltage and
(b) its internal resistance.

(a) 20.91 V

(b) 0.729 Ω

Explanation:

Using

E = I(R+r) Equation 1

Where E = Battery voltage, I = current, R = external resistance, r = internal resistance.

For the first case,

Given: I = 9.73 A, R = 1.42 Ω

Substitute into equation 1

E = 9.73(1.42+r)

E = 13.8166+9.73r Equation 2

For the second case,

Given: I = 7.36 A, R = 2.11 Ω

Substitute into equation 1

E = 7.36(2.11+r)

E = 15.5296+7.36r Equation 3

(a)

E = 13.8166+9.73r Equation 2

E = 15.5296+7.36r Equation 3

Make r the subject of the formula in equation 3

r = (E-15.5296)/7.36 Equation 4

Solving equation 2 and 3 simultaneously,

Substitute equation 4 into equation 2

E = 13.8166+9.73(E-15.5296)/7.36

E = 13.8166+1.3184(E-15.5296)

E = 13.8166+1.3184E-20.474

E-1.3184E = 13.8166-20.474

-0.3184E = -6.6574

E = -6.6574/-0.3184

E = 20.91 V.

Hence the  battery voltage = 20.91 V

(b)

Substitute the value of E into equation 2

20.91 = 13.8166+9.73r

9.73r = 20.91-13.8166

9.73r = 7.0934

r = 7.0934/9.73

r = 0.729 Ω

Hence the internal resistance = 0.729 Ω

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