Question

Why is this capacitor useful?

A 600`\mu`F capacitor is connected in parallel to a circuit containing a 5.0`\Omega` coil of resistance. A current of `2.0*10^3` A is passed during `1.4*10^(-3)`s

How do I explain why this capacitor is useful for providing the current for that amount of time?

Answer 1

`RC` time constant of the circuit `tau=600xx10^-6xx5.0=3\ m\ s`
Current passes for `1.4\ m\ s` which is roughly half of `tau`

It is given that a current of `2.0xx10^3\ A` is passed during `1.4xx10^-3\ s`.

The usefulness of this charged capacitor is to act like a voltage source to provide given current to the circuit the during given time interval as shown below.

.-.-.-.-.-.-.-.-.-.-.-

Capacitor `C` is connected in parallel to a circuit containing a coil of resistance `R` as shown in the figure. The capacitor is charged with initial charge `=Q_0`. Voltage across capacitor is equal to voltage across resistor.

`:.V_C=V_R`
`=>Q/C=iR`
where `i` is the current flowing.

`=>Q/C=-(dQ)/dtR`
`-ve` sign indicates that charge is decreasing

Reqriting and solving the differential equation we get for the discharge of capacitor

`(dQ)/dt=-1/(RC)Q`
`Q(t)=Q_0e^(-t/(RC))`

And for current in the circuit

`|i(t)|=|(dQ)/dt|=(Q_0/(RC))e^(-t/(RC))=i_0e^(-t/(RC))`

We see that the charge and current decay exponentially. Such heavy currents can be sustained only for a short duration which is shorter than `tau`.