In Figure 18.54, each capacitance C1 is 6.9μF and each capacitance C2 is 4.6μF , (a) Compute the

In Figure 18.54, each capacitance C1 is 6.9μF and each...

Key Concepts

Series and Parallel Capacitor Networks

This concept involves understanding how capacitors behave when connected in series or parallel. In series, the same charge passes through each capacitor but the voltages across them add up, whereas in parallel, each capacitor experiences the same voltage across its terminals but the charge stored can vary by capacitance. Recognizing these configurations is essential for simplifying complex circuits into effective equivalent capacitance values.

Equivalent Capacitance

Equivalent capacitance is the single capacitance that can replace a network of capacitors without altering the overall electrical behavior of the circuit. Calculating the equivalent capacitance involves using the rules for series (reciprocal sum formula) and parallel (direct sum) combinations. This concept is key for reducing complex networks to manageable simpler circuits that are easier to analyze.

Capacitor Charge-Voltage Relationship

This concept is based on the fundamental relationship Q = C × V, where Q is the charge stored on the capacitor, C is its capacitance, and V is the voltage across it. This relationship allows for calculating the amount of charge stored when a voltage is applied to the capacitor and is essential in analyzing energy storage and distribution in circuits that include capacitors.

Transcript

00:02 In the given problem there are few capacitors arranged in a mixed grouping between a and b.

00:11 First of all, here this is a capacitor having capacitance c1, another c1, one more c1, the fourth c1, then opposite to it again the same branch having three capacitors.

00:32 C1 capacitance each all of them are having a value of c1 then the two more capacitors but having a different value c2 yes here these two terminals are given as c and d but we consider ourselves these two terminals also as x and y.

01:12 Here this was also c2.

01:14 Now the values of this c1 and c2 are c1 is 6 .9 micro ferret and c2 is 4 .6 microferret in the first part of the problem.

01:29 We have to find the net capacitance between a and b.

01:36 So first of all, if you look carefully, there are 3 .5.

01:40 Capacitors having the capacitance 7 each which are in series like this so first of all we will find the net capacitance of these three capacitors net capacitance of three capacitors each having identical value will be let it be c s and that will be the identical value divided by the number of capacitors which is three so here it comes out to be 6 .9 divided by 3 and it becomes 2 .3 micro -farrid now this c s is in parallel with this c2 now the c s is in parallel with c2 so their combination the parallel combination will come out to be and we will let it be c cd the branch cd which will be cd is equal to c s plus c2 means this is 2 .3 plus 4 .6 micro ferret which again comes to to be 6 .9 micro ferret means if you look at the figure again this complete branch having four capacitors this is equivalent to 6 .9 microferred so again this c1 and this c1 these two are 6 .9 microferred and this one is also 6 .9 microferred hence we can see now 2c1 and this one c cd are again in series and identical in value so now the net capacitance and we will say to be c s dash the series combination but dash s dash again 6 .9 divided by 3 means 2 .3 micro ferret further this c s dash again is in parallel with this c2 and this time the parallel combination will be named as cxy so so cxy is equal to c s dash plus further c2 again this is 2 .3 plus 4 4 .6 which again comes out to be equal to 6 .9 micro -farrade so finally if i look again this c1 and this c1 both are having 6 .9 and this capacity net capacitance between x and y, this is again 6 .9.

05:02 So the net capacitance finally between a and b is equal to the series combination of 3 6 .9 microferret capacitors.

05:14 So it will be 6 .9 by 3...

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