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A current given by I(t)=4t4, where I is in amps when t is in seconds, passes through a 2-Ω resistor. How much charge passes through the resistor between t=3 s and t=5 s?

Respuesta :

xero099

Answer:

[tex]q = 2305.6\,C[/tex]

Explanation:

The charge passed during the interval of time is found by the integration the current function:

[tex]q = \int\limits^{5}_3 {4\cdot t^{4}} \, dt\\q= \frac{4}{5} \cdot t^{5}|_{3}^{5}\\q = \frac{4}{5} \cdot [(5)^{5}-(3)^{5}]\\q = 2305.6\,C[/tex]

akande212

Answer:

I(t) = dq/dt = 4t⁴

Integrating the above expression for q

q = 4/5t⁵

q = 4/5(5)⁵- 4/5(3)⁵

q = 2305.6 C

Explanation:

This problem involves the relationship between charge and current. Current is simply the amount of charge flowing per unit time and is given by dq/dt. That is the ratio of the amount of charge flowing per unit time to the time taken as the time taken for the charge to flow approaches zero (becomes very small).

From this relationship, to get q requires simply integrating the above expression for dq/dt within the time boundries of t = 3s and 5s.

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