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7 October, 05:30

What potential difference is measured across a 19.9 Ω load resistor when it is connected across a battery of emf 2.46 V and internal resistance 0.561 Ω? Answer in units of V.

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  1. 7 October, 05:52
    0
    2.388 V

    Explanation:

    Using

    E = I (R+r) ... Equation 1

    Where E = emf, I = current, R = External resistance, r = internal resistance

    I = E / (R+r) ... Equation 2

    Given: E = 2.46 V, R = 19.9 Ω, r = 0.561 Ω substitute into equation 2 to get the current

    I = 2.46 (19.9+0.561)

    I = 2.46/20.461

    I = 0.12 A.

    From Ohm's Law,

    V = IR ... Equation 3

    Where V = Potential difference across the 19.9 Ω resistor

    Substitute the value of I and R into equation 3

    V = 19.9 (0.12)

    V = 2.388 V
  2. 7 October, 05:58
    0
    The potential difference measured is 2.388 V.

    Explanation:

    E = I (R + r)

    E is the emf (electromotive force) of the battery = 2.46 V

    R is the resistance of the resistor = 19.9 ohms

    r is the internal resistance = 0.561 ohm

    I (current) = E / (R + r) = 2.46 / (19.9 + 0.561) = 2.46/20.461 = 0.12 A

    V (potential difference) = IR = 0.12 * 19.9 = 2.388 V
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