Consider the circuit shown in the diagram below. Before the switch is closed, bo
ID: 2260092 • Letter: C
Question
Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged.
A) Immediately after the switch is closed, what is the amount of current supplied by the battery?
B) Assuming the switch remains closed for a long time, which capacitor will be the first to reach 95% of its final charge level?
C) What is the time constant for charging this capacitor?
Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged. Immediately after the switch is closed, what is the amount of current supplied by the battery? Assuming the switch remains closed for a long time, which capacitor will be the first to reach 95% of its final charge level? What is the time constant for charging this capacitor?Explanation / Answer
You need to understand the behavior of an inductance L. Initially the current through an inductor is zero, and the current increases with time as voltage across the inductor/L; that is dI/dt = V/L.
A. The inductance initially passes zero current, so I = 10/20 = 0.5 A.
B. In steady state dI/dt = 0 so the voltage across the inductor = 0. Thus you can treat the inductor as a short circuit, and I = 10/(20||20) = 10/10 = 1 A.