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 resista

Question

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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Thu Thủy 4 weeks 2021-08-17T12:59:16+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-08-17T13:00:23+00:00

    Answer:

    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

    0
    2021-08-17T13:00:28+00:00

    Answer:

    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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