Question

1. after the switch has been closed for a very long time it is suddenly opened. What happens? Specifically explain what happens to the current in each of the resistors and the charge on the capacitors?

2. What is the current supplied by the battery the instant after the switch is closed? What is the current supplied by the battery after the switch has been closed for a very long time?

Answer 1

I'm very sorry to say that according to the HOMEWORKLIB RULES we are allowed to solve only first question.

solution : We consider an electric circuit in which a source of power (voltage), Resistor and Capacitor .

There are two cases as :

case 1) : When Resistor and Capacitor are in Series :

initially the charge on capacitor is zero but after switch  is closed for long time then the charge on capacitor is such that the charge equal to  voltage divided by capacitor (value) and the current through  resistor is zero . After that it suddenly opened then the charge from capacitor started to decrease and  there is some current in resistor.

case 2 ) : When Resistor and Capacitor are in Parallel : After switch  is closed for long time then the charge on capacitor is such that the charge equal to  voltage divided by capacitor (value) and the current through  resistor is i such that i*r= v. After that it suddenly opened then the charge from capacitor will constant  and   current in resistor is also same as it in closed.