Consider the circuit shown in Fig. 𝐏 25.71 . The emf source has negligible internal resistance.

Consider the circuit shown in Fig. 𝐏 25.71 . The emf source...

Consider the circuit shown in Fig. $\mathbf{P} 25.71 .$ The emf source has negligible internal resistance. The resistors have resistances $R_{1}=6.00 \Omega \quad$ and $\quad R_{2}=4.00 \Omega$ The capacitor has capacitance $C=9.00 \mu \mathrm{F}$. When the capacitor is fully charged, the magnitude of the

Key Concepts

Thevenin's Theorem in Circuit Analysis

Thevenin's Theorem provides a method to simplify a complex portion of a circuit by replacing it with an equivalent circuit composed of a single voltage source and a single series resistor. This simplification is particularly useful for analyzing the behavior of capacitors within a circuit, as it allows for easier calculation of voltages and assessment of the capacitor's impact.

Voltage Division in Resistive Networks

When resistors are arranged in a circuit, the total voltage supplied by the source is divided among them based on their resistance values. Understanding voltage division is vital for determining the potentials at various points in the circuit, especially when some branches, like those containing a fully charged capacitor, are effectively inactive.

Steady-State Behavior of Capacitors in DC Circuits

In a DC circuit, after a long period the capacitor becomes fully charged and reaches a steady state. At this point, it effectively behaves as an open circuit and no longer allows current to flow through its branch, which is a critical factor when analyzing the overall circuit behavior.

Open-Circuit Characteristics of a Fully Charged Capacitor

A fully charged capacitor establishes a constant voltage across its plates and then stops drawing any current. This open-circuit condition plays a crucial role in determining how the remaining circuit elements interact, as the capacitor essentially isolates the parts of the circuit it is connected to.

Transcript

00:01 Hi, here in the given circuit, this is the resistance r1, then there is a capacitor, having the capacitance c, which is having another resistance r2 in its parallel, and then the battery having emf epsilon.

00:30 Suppose these terminals are named as a, b.

00:37 Internal resistance of the battery is negligible, let it be zero.

00:42 The values of resistances are r1 is 6 .00 oom, r2 is 4 .00 oom.

00:54 And the value of capacitor is c, means capacitance is c is equal to 9.

01:00 0 .00 micro ferret.

01:05 Now when the capacitor is fully charged means steady state, no current will be passing through the branch having capacitor.

01:39 Whole of the current will be passing through these two resistors which will serve as to be in series combination.

01:46 Whatever the current there will be that will be passing through this series combination of the two resistors.

01:53 So it is given that the charge is stored over the capacitor in steady state is 36 .0 microculum.

02:05 So as we know the value of capacitance, we know the value of charge stored over it.

02:10 Hence the potential drop across the plates of the capacitor, let it be vab, that will be given by v is equal to q by c.

02:22 For q, this is 36 .0 microculum.

02:27 For c, this is 9 .00 micro -farrid.

02:31 So this value of the potential drop across ab comes out to be 4 .00 volt as potential remains the same in parallel.

02:43 So the resistance r2 will also be having same potential drop across 8...

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