The capacitor in an R C circuit with a time constant of 15...

The capacitor in an $R C$ circuit with a time constant of $15 \mathrm{ms}$ is charged to $10 \mathrm{V}$. The capacitor begins to discharge at $t=0 \mathrm{s}$. a. At what time will the charge on the capacitor be reduced to half its initial value? b. At what time will the energy stored in the capacitor be reduced to half its initial value?

The capacitor in an $R C$ circuit with a time constant of $15 \mathrm{ms}$ is charged to $10 \mathrm{V}$. The capacitor begins to discharge at $t=0 \mathrm{s}$.
a. At what time will the charge on the capacitor be reduced to half its initial value?
b. At what time will the energy stored in the capacitor be reduced to half its initial value?

Key Concepts

RC Circuit

An RC circuit is an electric circuit composed of resistors and capacitors, where the capacitor charges or discharges through the resistor over time. This type of circuit is fundamental in studying transient responses, filtering signals, and timing applications in electronics.

Time Constant

The time constant, denoted by ? (tau), is the product of resistance (R) and capacitance (C). It characterizes the rate at which the voltage across the capacitor rises during charging or falls during discharging, with the voltage changing by approximately 63% of the difference between initial and final values in one time constant.

Exponential Decay

Exponential decay in the context of an RC circuit refers to the mathematical modeling of the voltage and charge decrease over time. The magnitude diminishes exponentially, meaning that it drops rapidly at first and then more slowly as time increases. This behavior is governed by the equation V(t) = V(0) * exp(-t/?).

Stored Energy in a Capacitor

The stored energy in a capacitor is determined by the formula E = ½ CV². Since the voltage decays exponentially during discharge, the energy stored, which depends on the square of the voltage, also decreases exponentially but with a different decay rate. Understanding this relationship is crucial when considering energy reduction in transient circuit analysis.

Transcript

00:01 In the given problem, time constant of an rc circuit is given as tau is equal to 15 milliseconds.

00:17 This capacitor, the capacitor put in this circuit has been charged up to a potential of 10 volt.

00:24 Now, in the first part of the problem, we have to find the time taken by the charge is stored over the capacitor to decay.

00:34 Up to half of its initial value means q is equal to q0 by 2.

00:40 So using the equation of discharging of a capacitor, which is given as q is equal to q0 into e -r -to -the -power minus t by tau.

01:07 Here this cue is given as q -0 by 2 is equal to q -0 into e -rate -to -per -minus t by tau.

01:14 Canceling this q0, we get e raised to the power minus t by 15 milliseconds is equal to 0 .5.

01:27 So taking log base e at both of the sides, it becomes log at base e of base e of e of e millisecond is equal to log base e of 0 .5.

01:50 Using rules of logarithm here, this comes out to be 1.

01:55 So it remains as minus t by 15 milliseconds is equal to the value of this log base e 0 .5 comes out to be minus 0 .693.

02:07 So canceling this negative sign here t comes out to be 0 .693 into 15 milliseconds or we can say this time is actually equal to 10 .4 millisecond means in a time of 10 .4 milliseconds the charge is stored over the capacitor will be reduced to half of its initial value.

02:34 Now in the second part of the problem we have to find the time taken by the capacitor to reduce its energy, the energy is stored over it to half of its initial value means u is equal to u0 by 2.

02:49 Now using the expression for energy as u is equal to half c v square as there will be no change in the capacity tense, there will be change only in the potential.

03:02 So for you this is half cv square is equal to for u not this is half cv .s...

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