In Fig. 𝐄 2 4 . 2 0, C1= 6.00 μF, C2=3.00 μF, and C3= 5.00 μF. The capacitor network is connec

step 1 Hi, here in this given problem there is a grouping of capacitors. Two capacitors are in parallel

In Fig. 𝐄 2 4 . 2 0, C1= 6.00 μF, C2=3.00 μF, and C3= 5.00...

$\mathrm{In}$ Fig. $\mathbf{E 2 4 . 2 0}, \quad C_{1}=$ $6.00 \mu \mathrm{F}, C_{2}=3.00 \mu \mathrm{F},$ and $C_{3}=$ $5.00 \mu \mathrm{F}$. 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 $30.0 \mu \mathrm{C}$. (a) What are the charges on capacitors $C_{1}$ and $C_{3}$ ? (b) What is the applied voltage $V_{a b} ?$

$\mathrm{In}$ Fig. $\mathbf{E 2 4 . 2 0}, \quad C_{1}=$
$6.00 \mu \mathrm{F}, C_{2}=3.00 \mu \mathrm{F},$ and $C_{3}=$
$5.00 \mu \mathrm{F}$. 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 $30.0 \mu \mathrm{C}$.
(a) What are the charges on capacitors $C_{1}$ and $C_{3}$ ?
(b) What is the applied voltage $V_{a b} ?$

Key Concepts

Voltage Division in Capacitor Circuits

Voltage division in capacitor circuits, especially in series configurations, refers to the way in which an applied voltage is divided among the capacitors. The voltage across each capacitor is determined by its capacitance relative to the sum of the capacitances in series. This principle is essential for determining the voltage across individual components in the network when the total voltage is known.

Charge Distribution in Capacitor Networks

Analyzing the distribution of charge in a network of capacitors involves using the properties of the specific configuration (series or parallel). In series configurations, the same charge appears on each capacitor, whereas in parallel configurations, the charge on each capacitor depends directly on its capacitance. This concept is key to finding unknown charges and voltages in complex circuits.

Series and Parallel Capacitor Configurations

Capacitors can be connected in series or parallel, and each configuration affects the overall behavior of the circuit differently. When connected in series, the same charge flows through each capacitor, while the voltage divides among them inversely proportional to their capacitance. In parallel, each capacitor experiences the same voltage, and the overall capacitance is the sum of the individual capacitances.

Capacitors and Their Behavior

Capacitors are devices that store electrical energy through the accumulation of electric charge. They are characterized by the relationship Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. Understanding this relationship is fundamental when analyzing how charges and voltages distribute in a circuit.

Equivalent Capacitance

The equivalent capacitance is a single capacitance value that can replace a network of capacitors without changing the overall electrical behavior of the circuit. Calculating the equivalent capacitance is crucial for simplifying the analysis and determining how the entire network responds to an applied voltage.

Transcript

00:01 Hi, here in this given problem there is a grouping of capacitors.

00:08 Two capacitors are in parallel.

00:15 This is c1 and c2.

00:22 Terminal a then attached with the terminal b, there is third capacitor c3.

00:32 The values of these capacitors, capacitors, c1.

00:37 This is 6 .00 micro -farrate, c2, that is 3.

00:43 0 .00 micro ferret and c3 this is 5 .00 micro ferret charge over the capacitor c2 q2 that is given as 30 .0 microculum in the first part of the problem we have to find the charge over other two capacitors means q1 we have to find and q3 we have to find first of all we find potential across this c2, potential drop across c2, that is v2 is equal to q2 by c2, means 30 .0 microculum divided by c2 which is 3 .00 micro -farrad and it is calculated to be equal to 10 .0 volt and as we know potential remains same in parallel.

01:41 So this will be equal to v1 also.

01:45 As potential drop remains the same in parallel.

01:58 So for c1 now we may find q1, q1 will be equal to c1 multiplied by v1 means this is 6 .00 microferred multiplied by 10 volt which we have just found.

02:14 So this q1 will come out to be equal to 60 .0 microculam, one of the answer.

02:21 Now q3 that will be equal to q1 plus q2, the total charge q1 plus q2.

02:31 This is 60 .0 plus 30 .0 microculeum means this is 90 .0 microculem.

02:43 So the two answers for the first part of the problem.

02:49 Q1 that is 60 .0 microculum and q3 that is 90 .0 microculum...

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