Question

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 resistor at t = 1.0 s, after the switch is closed is: a. 64 mA
b.25 mA
c.49 mA
d.64 mA

Answer 1

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.

Answer 2

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