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 w

Question

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.

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Minh Khuê 4 years 2021-08-14T09:54:27+00:00 1 Answers 8 views 0

Answers ( )

  1. Maris
    0
    2021-08-14T09:56:22+00:00
    Reply

    Answer:

    (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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