(III) Two resistors and two uncharged capacitors are...

(III) Two resistors and two uncharged capacitors are arranged as shown in Fig. $19-58$ . Then a potential difference of 24 $\mathrm{V}$ is applied across the combination as shown. (a) What is the potential at point a with switch S open? (Let $V=0$ at the negative terminal of the source. (b) What is the potential at point b with the switch open? (c) When the switch is closed, what is the final potential of point b? (d) How much charge flows through the switch S after it is closed?

Key Concepts

Voltage Divider

A voltage divider is a fundamental circuit where two or more resistors in series divide an input voltage into smaller voltages at their junctions. This principle is used to determine the potential at intermediate points in a circuit, making it a key tool in circuit analysis.

RC Circuit Behavior

An RC circuit consists of resistors and capacitors and exhibits time-dependent behavior. When such circuits are subject to a voltage source, the capacitor charging process involves an initial transient period followed by a steady state where the capacitor behaves like an open circuit. This concept is crucial for analyzing the evolution of voltage and charge distribution in circuits over time.

Capacitor Charging and Discharging

Capacitors store electrical energy in the form of charge and voltage. They charge through resistors exponentially over time when connected to a voltage source, and when discharged they also release their stored energy exponentially. This behavior is described by time constants and is central to understanding the dynamics of circuits that include capacitors.

Charge Redistribution

When circuit configurations change, for example by closing a switch, the stored charge on capacitors may redistribute to reach a new equilibrium state. This concept involves the conservation of charge and energy, and is essential in predicting the final voltages and the amount of charge transferred in dynamic circuit systems.

Transcript

00:02 In the first part of this problem, we want to find potential at point a when switch is open.

00:14 So when switch is open, then the two registers are in series with the battery.

00:29 We can find the amount of current passes through the two registers by using kitchov's voltage rule.

00:50 That is 24 volt minus i into 8 point and om minus i into 4 .4 om is equal to 0 this gives us i equal to 1 .8 .1 at ampere.

01:28 So this is the current passes through both registers that is 1 .8 .m.

01:43 Now we can easily calculate the potential difference drop across the second register that is the resistor having resistance of 4 .4 om by using oms allow we have voltage drop equal to current into resistance here current is 1 .8 1 ampere into resistance is 4 .4 om this gives us the voltage drop across the register equal to 8 .0 volt as it is mentioned in the problem that negative terminal of battery is taken as v equal to 0 and we can easily tell that potential at point a is equal to 8 .0 volt now in the second part of this problem we want to find potential at point b when switch is open as when switch is open then both the capacitors are connected in series with the battery of emf 24 volt first we will calculate net capacitance net capacitance in series combination equal to c1 into c2 divided by c1 plus c2 capacitance of first capacitor is 0 .48 microfarrat into capacitance of second capacitor is 0 .24 microfaird divided by 0 .48 microfarrat plus 0 .4 24 micro ferret this gives us net capacitance equal to 0 .6 0 .16 micro ferret as in series combination charged stored by first capacitor is equal to charged stored by second capacitor is equal to charge stored by the equivalent capacitor and this is equal to capacitance of equivalent capacitor into voltage of battery as capacitance of equivalent capacitor is 0 .16 micro ferret that is 0 .16 times 10 to the power minus 6th a heart into voltage of battery is 24 volt this gives us the charge stored by net capacitor equal to 3 .4it micro column...

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