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vagabundo [1.1K]
2 years ago
13

A 4.8 mF capacitor in series with a 500 Ω resistor is connected, by a switch, to a 12 V battery. The current through the resisto

r at t = 1.0 s, after the switch is closed is: a. 64 mA
b.25 mA
c.49 mA
d.64 mA
Physics
2 answers:
bonufazy [111]2 years ago
4 0

Answer:

The current is 15.8 mA.

Explanation:

Given that,

Capacitor = 4.8 mF

Resistor = 500 Ω

Time t = 1.0 s

Voltage = 12 V

We need to calculate the current

Using formula of current

I(t)=\dfrac{V_{0}}{R}e^{\dfrac{-t}{RC}}

Where, V = voltage

R = resistance

t = times

Put the value into the formula

I(t)=\dfrac{12}{500}e^{\dfrac{-1.0}{500\times4.8\times10^{-3}}}

I(t)=0.0158\ A

I(t)=15.8\ mA

Hence, The current is 15.8 mA.

ad-work [718]2 years ago
3 0

Answer:

Current will be 81.7 mA

Which is not given in bellow option

Explanation:

We have given  capacitance C=4.8mF=4.8\times 10^{-3}F

Resistance R = 500 ohm

Voltage V = 12 volt

We know that time constant of RC circuit of RC circuit is given by

\tau =RC=500\times 4.8\times 10^{-3}=2.4sec

Time is given as t = 1 sec

We know that current in RC circuit is given by

i=\frac{v}{R}(1-e^{\frac{-t}{\tau }})

So current i=\frac{12}{500}(1-e^{\frac{-1}{2.4 }})=0.00817A=81.7mA

Which is not given in the following option

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Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

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