Question

Consider a RC circuit. At time t=0, the circuit is closed. ( a) Draw how the current behaves with time. (b) What about the power dissipated by the resistor ? (also draw it as a function of t)

Answer 1

Kirchhoff’s Loop Rule

At t=0, there is no charge on (hence no potential across) the capacitor so at the “instant” the switch is closed, we expect a current of I0=ε/R (voltage drop is only across the resistor). As the capacitor charges, we expect the current to fall.

   

                     

Close Switch at t=0

Charge will flow on the capacitor and the current will decrease until we reach i = 0, then:

                 

the final value charge, Q_f does not depend on R

     

                                     

                                 

     

This is a differential equation. We can rearrange and integrate to find what we need.

                                     

                                   

       

                                

      Taking the exponents (inverse log) both sides

                              

                         

                          

Taking the time derivative, to get the instantaneous , current, i  

                           

                            

                      

Solution(B)

            

              

                        

   

  

Kirchhoff’s Loop Rule

At t=0, there is no charge on (hence no potential across) the capacitor so at the “instant” the switch is closed, we expect a current of I0=ε/R (voltage drop is only across the resistor). As the capacitor charges, we expect the current to fall.

   

                     

Close Switch at t=0

Charge will flow on the capacitor and the current will decrease until we reach i = 0, then:

                 

the final value charge, Q_f does not depend on R

     

                                     

                                 

     

This is a differential equation. We can rearrange and integrate to find what we need.

                                     

                                   

       

                                

      Taking the exponents (inverse log) both sides

                              

                         

                          

Taking the time derivative, to get the instantaneous , current, i  

                           

                            

                      

Solution(B)