24.18. In Fig. 24.26, C1= 6.00 μF, C2=3.00 μF, and C3=5.00 μF . The capacitor network is con

24.18. In Fig. 24.26, C1= 6.00 μF, C2=3.00 μF, and...

24.18. In Fig. $24.26, \quad C_{1}=$ $6.00 \mu F, \quad C_{2}=3.00 \mu F,$ and $C_{3}=5.00 \mu F .$ The capacitor network is connected to an applied potential $V_{a b}$ . After the charges on the capacitors have reached their final values, the charge on $C_{2}$ is 40.0$\mu \mathrm{C}$ . (a) What are the charges on capacitors $C_{1}$ and $C_{3} ?(b)$ What is the applied voltage $V_{a b} ?$

Key Concepts

Equivalent Capacitance Calculation

When multiple capacitors are connected together, the concept of equivalent capacitance is used to simplify the analysis of the circuit. In parallel, equivalent capacitance is the sum of the individual capacitances, and in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. This simplification enables easier calculations of the overall charge and voltage distributions in the circuit.

Voltage and Charge Distribution in Networks

By using the principles of voltage division in series circuits and equal voltage across capacitors in parallel, as well as the conservation of charge, one can determine the individual charges and applied potential in a complex capacitor network. This concept involves understanding how the initial applied voltage is partitioned across various elements and how the network reaches an equilibrium state with shared charges and voltages.

Capacitor Q=CV Relationship

This relationship, expressed as Q = C × V, is fundamental in understanding how capacitors store energy. It explains that the charge (Q) stored on a capacitor is directly proportional to its capacitance (C) and the voltage (V) across its plates, thus serving as a primary equation when analyzing circuits that include capacitors.

Series and Parallel Capacitor Configurations

Capacitors in a circuit can be connected in series or parallel, and each configuration has distinct rules for how charge and voltage are distributed among the components. In series configurations, the same charge flows through each capacitor while the voltages add up, whereas in parallel configurations, each capacitor experiences the same voltage while the individual charges may vary depending on their capacitances.

Transcript

00:01 The circuit diagram of a system of capacitors is as shown in the given problem.

00:11 This is point a after which there is a parallel combination of two capacitors, c1 and c2.

00:22 This is c2.

00:26 Which then connected with a point d which is further connected with a point d which is further connected with third capacitor in series this is c3 and finally this is a point b the values given over here are c1 is given as 6 .00 micro ferret for c2 it is 3 .00 micro ferret and for c3 this is 5 .00 microcarat now the charge given over this capacitor c2.

01:07 The value of charge is q2 is equal to 40 .0 microculum.

01:18 In first part of the problem we have to find the charges on the other two capacitors c1 and c3.

01:25 To do this first of all we will find the potential drop across the plates of this capacitor c2 so this potential across this capacitor is found the formula v2 is equal to q2 by c2 so here q2 is this 40 .0 microculum and c2 is 3 .0 micro so this potential can be given as 40 by 3 volt as these two capacitors c1 and c2 are in parallel and we know potential remains the same in parallel as potential remains same in a parallel combination so we can say v1 will also be equal to v2 means 40 by 3 volt so finally we can find the value of charge over this capacitor of capacitance c1 so q1 is equal to c1 multiplied by v1 and c1 was 6 .00 micro fared multiplied by v1 which is nothing but 40 by 3 volt so the value of q1 here comes out to be 80 .0 microcolon this is first answer of the first part of the problem as we have to find the charges on the other two capacitors so we found the charge on the first capacitor c1 this is 80 microculum now total charge in this parallel combination q will be actually equal to q3 because when the charge will be combined over here in this parallel combination and it will become q so this is actually the charge which is passing through this third capacitor so q net q will be equal to q3 so q is equal to q3 is equal to q3 is equal to q1 plus q2 or we can say this is 40 microculum plus 80 microculem or we can say this is q3 is equal to 120 microculum so this becomes the answer for the second part of the first part of the problem finally we have to find so these are the two answers now we have to find the net potential difference between these two points...

Please give Ace some feedback