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?

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.

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en.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#.