How long will it take for the current in a circuit to drop...

How long will it take for the current in a circuit to drop from its initial value to $1.50 \mathrm{~mA}$ if the circuit contains two $3.80-\mu \mathrm{F}$ capacitors that are initially uncharged, two $2.20-\mathrm{k} \Omega$ resistors, and a 12.0 - $\mathrm{V}$ battery all connected in series?

Key Concepts

RC Circuit

An RC circuit is one that contains both resistors and capacitors. The interplay between the resistor and the capacitor defines the transient behavior of the circuit, such as how fast it charges or discharges, which is central to analyzing time-dependent voltage and current changes.

Time Constant

The time constant of an RC circuit, typically denoted ? (tau), is the product of the resistance and the capacitance in the circuit. It characterizes the rate at which the circuit responds to changes: after a time equal to one time constant, the voltage across the capacitor, or the current in the circuit, will have reached approximately 63% of its final value in charging or decayed to 37% in discharging.

Exponential Decay

Exponential decay in RC circuits describes how the current or voltage changes over time, usually following a mathematical form such as I(t) = I0 e^(–t/?). This concept is crucial for understanding transient analysis since it shows that the rate of change is proportional to the amount remaining to be changed, leading to a smooth, continuous decay over time.

Equivalent Capacitance in Series

When capacitors are connected in series, the overall or equivalent capacitance is less than any individual capacitor's value. This is determined by the reciprocal sum of the individual capacitances, and it is important to calculate this equivalent capacitance to accurately compute the overall time constant of the circuit.

Transcript

0:00 Hi.

00:02 In the given problem, initial current in cr circuit is given as i .0 is equal to v by r s.

00:30 As here in the given circuit two resistors are in series combination, so net resistance is r s.

00:37 And the initial current means when the capacitors starts getting charged.

00:46 Initially there was no charge over the capacitors and initial current will be this the maximum current this will be.

00:54 Now as the capacitors start getting charged this current starts decreasing so instantaneous current in the circuit is given as i is equal to 1 .50 millil ampere and we have to find the time in which this current i0 will drop to i now the capacitance of each capacitor is given as c is equal to 3 .80 micro ferret and as there are two capacitors in series so net capacitance in this series combination will be c by 2 3 .80 by 2 means this is 1 .90 micro ferret in the similar manner resistance of each resistor out of the two resistors which are in series that is given as 2 .20 kilo on so net resistance of the circuit r s that will begin by 2 into r or we can say this is 2 into 2 .20 kilo -oam means 4 .40 kilo -oam this is the net resistance so time constant of this cr circuit will be tau is equal to the product of net capacitance with the net resistance means this is 1 .90 micro ferret or 1 .90 into 10 dash to per minus 6 ferret multiplied by 4 .40 kilo ome or 4 .40 into 10 dash per 3 and it comes out to be 8 .36 milliseconds.

03:58 So now using the equation of decreasing current in a charging capacitor and that equation says instantaneous current i is equal to maximum current i0 into e -resd power minus t by tau.

04:35 So, plugging in all the known values here, this is i is equal to for i nor this will be v by r s into e -r -r -r -m minus t by tau.

04:49 Now we will put the known values.

04:52 For i this is 1 .50 millie -amper is equal to for v 12 volt divided by net resistance 4 .40 into 10 -rish to par 3...

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