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

The current is 15.8 mA.

Step-by-step 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)=(V_(0))/(R)e^{(-t)/(RC)}`

Where, V = voltage

R = resistance

t = times

Put the value into the formula

`I(t)=(12)/(500)e^{(-1.0)/(500*4.8*10^(-3))}`

`I(t)=0.0158\ A`

`I(t)=15.8\ mA`

Hence, The current is 15.8 mA.

Answer 2

Current will be 81.7 mA

Which is not given in bellow option

Step-by-step explanation:

We have given capacitance
`C=4.8mF=4.8* 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* 4.8* 10^(-3)=2.4sec`

Time is given as t = 1 sec

We know that current in RC circuit is given by

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

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

Which is not given in the following option