3 years ago

Evalaute (1/2^-2) + 1/2

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

I hope this helps you

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3 years ago

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin

hoa [83]

Answer:

$\frac{dI}{dt}=-0.00016 A/s$

Step-by-step explanation:

We are given that

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$V=IR$

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I=0.03 A

$\frac{dV}{dt}=-0.06V/s$

$\frac{dR}{dt}=0.07ohm/s$

We have to find rate of change of current I means $\frac{dI}{dt}$

Differentiate the equation w.r.t t

$\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}$

Substitute the values then we get

$-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07$

$-0.06=389\frac{dI}{dt}+0.0021$

$-0.06-0.0021=389\frac{dI}{dt}$

$-0.0621=389\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s$

Hence, the current I is changing at the rate=-0.00016A/s

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4 years ago