Question

In the circuit below, R1 = 5.5 Ohms, R2 = 4.6 Ohms, and C = 4.5 micro-farads. The capacitor was initially charged by a battery and right when the battery was disconnected, the charge in the capacitor is 215.2 micro-coulombs. What is the current through R1, in amps, 15.4 micro-seconds after the battery has been disconnected? Hint: combine the resistors into their equivalent form and use this to find the time constant of the equivalent circuit. Then, find the voltage across R1 and use the fact that everything here is connected in parallel.

Answer 1

When the battery is removed, the dielectric will decrease the electric field strength and the voltage between the plates while it increases their capacitance. Diagram on the Circuit not given Explanation can be found Before the switch is closed, there is no charge on the capacitor, so the voltage is zero across the capacitor at this time. Because it is not possible to change the charge on the capacitor like a step function (or the current should be infinitely large), immediately after the switch is closed, the voltage across the capacitor (and R2 ) is still zero. Therefore, the voltage across R 1 is V; i.e. think of the capacitor as being a short-circuit for this instant of time. So the current supplied by the battery, which is the same as the current going through R1 , is I 0 E = V/d Using the fact that V = Ed and that capacitance is the ratio of the charge stored per unit volt we derived the following formula for the capacitance based on the geometry of a parallel-plate capacitor.

Refer to the following information for the next five questions. A 90