Question

In the circuit in the figurethe capacitors are all initiallyuncharged, the battery has no internal resistance, and the ammeteris idealized.Part AFind the reading of the ammeter just afterthe switch S is closed.Part BFind the reading of the ammeter after theswitch has been closed for a very long time.

Answer 1

IDENTIFY: Inboth cases, simplify the complicated circuit by eliminating theappropriate circuit elements. The

potentialacross an uncharged capacitor is initially zero, so it behaves likea short circuit. A fully charged capacitor

allows nocurrent to flow through it.

(a)SET UP: Justafter closing the switch, the uncharged capacitors all behave likeshort circuits, so any resistors

in parallelwith them are eliminated from the circuit.

EXECUTE: Theequivalent circuit consists of 50 Ω and 25 Ω in parallel, with this combination inseries with

75Ω, 15 Ω, and the 100-V battery. Theequivalent resistance is 90 Ω +16.7 Ω= 106.7 Ω, which gives

I = (100V)/(106.7 Ω) = 0.937A.

(b)SET UP: Longafter closing the switch, the capacitors are essentially charged upand behave like open circuits

since nocharge can flow through them. They effectively eliminate anyresistors in series with them since no

current canflow through these resistors.

EXECUTE: Theequivalent circuit consists of resistances of 75 Ω, 15 Ω, and three 25-Ω resistors, all in serieswith

the 100-Vbattery, for a total resistance of 165 Ω. Therefore I = (100 V)/(165 Ω) = 0.606 A

EVALUATE: Theinitial and final behavior of the circuit can be calculated quiteeasily using simple series-parallel

circuit analysis.Intermediate times would require much more difficultcalculations!pls vote as soon asposible...