Why is this capacitor useful?

A 600\muF 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?

1 Answer
Jun 4, 2018

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.

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

en.wikipedia.orgen.wikipedia.org

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.