A capacitor with C=1.50 ×10^-5 F is connected as shown in Fig. 26.61 with a resistor with R=980

A capacitor with C=1.50 ×10^-5 F is connected as shown in...

A capacitor with $C=1.50 \times 10^{-5} \mathrm{F}$ is connected as shown in Fig. 26.61 with a resistor with $R=980 \Omega$ and an emf source with $\mathcal{E}=18.0 \mathrm{V}$ and negligible internal resistance. Initially the capacitor is uncharged and the switch $\mathrm{S}$ is in position $1 .$ The switch is then moved to position $2,$ so that the capacitor begins to charge. After the switch has been in position 2 for 10.0 $\mathrm{ms}$ , the switch is thrown from position 2 back to position 1. (b) Compute the voltage drops across the resistor and across the capacitor at the instant described in part (a). (c) Compute the voltage drops across the resistor and across the capacitor just after the switch is thrown from position 2 back to position 1 . (d) Compute the charge on the capacitor 10.0 $\mathrm{ms}$ after the switch is thrown from position 2 back to position $1 .$

Key Concepts

RC Circuit

An RC circuit involves a resistor (R) and a capacitor (C) connected together, where the resistor controls the rate at which the capacitor charges or discharges. The behavior of such circuits is characterized by the interplay between the resistor’s dissipative effects and the capacitor’s ability to store and release energy, leading to time-dependent voltage and current changes.

Time Constant

The time constant, represented by the product RC, is a crucial parameter in RC circuits that determines the rate at which the capacitor charges or discharges. It provides a measure of the time required for the capacitor to charge up to about 63% of its final voltage or discharge to about 37% of its initial voltage, thereby governing the exponential rate of change in the circuit.

Exponential Charging and Discharging

The process of charging a capacitor in an RC circuit occurs exponentially, following the equation Vc = V(1 - e^(-t/RC)), where Vc is the capacitor voltage at time t and V is the supply voltage. Similarly, when discharging, the capacitor voltage decreases exponentially. This exponential behavior arises from the differential equation that governs the current and voltage relationships in the circuit.

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) states that the sum of the electric potential differences around any closed loop in a circuit must be zero. This principle is used to relate the voltage drops across the resistor and capacitor with the supply voltage, particularly during the transient phases when the circuit is switching between different configurations.

Transient Analysis

Transient analysis involves studying the behavior of electrical circuits as they respond to changes, such as when a switch is thrown. During these transient events, the voltages and currents in the circuit adjust from one steady state to another, and understanding this behavior requires solving the governing differential equations that incorporate initial conditions and the time constant of the RC circuit.

Transcript

00:01 Hi, in the given problem, this is a circuit diagram which is having a switch s, which is having two possible positions, one and two.

00:15 When in position one, the circuit is having simply a capacitor and a resistor without any source of him.

00:28 And then it is in position 2 it will be with a source of emm this is resistance r this is capacitance c the values given here are for c for capacitance this is 1 .50 into 10 dash to power minus 5 ferret the value of this resistance is 980 oom and the emf applied is 18 .0 volt.

01:05 Initially the switch was in position 1, so the capacitor was uncharged.

01:11 After that, it is thrown to situation 2, to the position 2.

01:15 So the capacitor starts getting charged.

01:20 Just after switching it to position 2, after 10 milliseconds in a time of just 10, milliseconds after switching it on from 1 to 2 the switch is thrown back to position 1 so in the first part of the problem we have to find the charge on the capacitor just before the switch is thrown from position 2 back to position 1 means to say simply we have to find the charge on the capacitor plates in a time of 10 milliseconds using the equation of charging capacitor so the equation of charging capacitor the equation of charging capacitor the equation for charge is given as q is equal to q not the maximum possible charge is q not over the plates of the capacitor 1 minus e raised the power minus t any instant t divided by rc which is the time constant of the cr circuit here, the time is given as 10 milliseconds or we can say this is 10 dash to power minus 2 seconds.

02:37 For the maximum possible charge, which may be accumulated over the plates of the capacitor, this is simply c into e, which comes out to be for c, this is 1 .50 into 10 dash bar minus 5 farrid multiplied by p which is 18 volt and this value of maximum possible charge comes out to be 13 .3 into 10 dash per minus 5 coularn this is the maximum possible charge so for a time of 10 dash per minus 2 second putting the values here in this equation number 1 we get the value of charge q to be equal to for q not this is 13 .3 into 10 dash per minus 5 1 minus e to the power minus 10 dash to power minus 2 divided by for r 9 80 for c this is 1 .5 into 10 dash power minus 5 so then we solve these values it comes out to be 13 .3 into 10 dash bar minus 5 so when we solve these values it comes out to be 13 .3 into 10 ratio per minus 5, 1 minus e ratio to power, it becomes minus 0 .68.

04:02 If we calculate it separately, this power of e comes out to be 0 .68 in magnitude.

04:10 So when we find the value of this e ratio minus 0 .68, this value here comes out to be this is 13 .3 into 10 to 10 x5, 1 minus and the value of this e ratio to the power minus 0 .6.

04:28 6 comes out to be 0 .5 .066.

04:34 So it becomes finally we can say this is discharge.

04:47 When we solve this it comes out to be 133 into 10 to the power minus 6 coulon or 133 microculum.

05:05 This is the answer for the first part of the problem the charge stored over the plates of the capacitor in this much time.

05:19 But wait here we have made a mistake the product of this 1 .5 and 18 will not be 13 .3 this is wrong here actually this value here will come out to be 27 this is 27 so here also it will be 27 it will be 27 here also then here but when we multiply it by the difference of one with 0 .5066 now it will come out to be 13 .3 so making a correction that this is 27 this is also 27 into 10 minus 5 and here it is also 27 which finally comes out to be 13 3 into 10 dashpar minus 5 coulum or 133 into 10xma minus 6 coulum or we can say this.

06:22 So finally this is the answer for the first part of the problem.

06:26 Now in the second part of the problem we have to find the voltage drop across the resistor and across the capacitor at this instant of 10 milliseconds...

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