A battery with an emf of 4 V and an internal resistance of 0.7 capital omega is connected to a variable resistance R. Find the current and p

Question

A battery with an emf of 4 V and an internal resistance of 0.7 capital omega is connected to a variable resistance R. Find the current and power delivered by the battery when R is (a) 0, I = 5.714285714 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 5.714285714 OK P = 0 W * [1.25 points] 3 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK (b) 5 capital omega, I = 0.701754386 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.701754386 OK P = 2.462296091 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 2.462296091 OK (c) 10 capital omega, and I = 0.3738317757 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.3738317757 OK P = 1.397501965 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 1.397501965 OK (d) infinite. I = 0 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK P = W

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Farah 3 years 2021-09-03T03:45:04+00:00 1 Answers 8 views 0

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  1. Sapo1
    0
    2021-09-03T03:46:19+00:00
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    Answer:

    E = I(R + r)

    Making I the subject of the formular by dividing both sides by R + r,

    I = E/(R + r)

    E = 4V, r = 0.7Ohm, R = 0

    I = 4/(0 + 0.7) = 4/0.7

    I = 5.174285714A

    Explanation:

    For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:

    I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms

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