Potential difference across ab in the network shown is

For 2 capacitors in series: Ct = C1 * C2/(C1 + C2) Ct = 18000/270 Ct = 66.6666666 nF A) Q = CV = 66.66666 nF * 53 V = 3,533.3333 nC = 3.5333333 uC B) The charge on all series connected caps will be the same. 3.5333333 uC C) Ditto 3.53333333 uC D) W = (1/2)CV² = (1/2) * 66.66666 nF * (53)² = 93,633.33 nJ = 93.6333 uJ E) First we need to know the voltage across the 150 nF cap. ...... V = Q/C = 3,533.3333 nC/150 nF = 23.55555 V ...... W = (1/2)CV² = (1/2) * 150 nF * (23.55555)² = 41,614.8 nJ = 41.615 uJ F) Voltage across the 120 nF ..... V = Q/C = 3533.3333 nC/120 nF = 29.44444 V ..... W = (1/2)CV² = (1/2) * 120 nF * (29.44444)² = 52,018.5 nJ = 52.010 uJ G) See E) H) See F

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)%2F08%253A_Capacitance%2F8.03%253A_Capacitors_in_Series_and_in_Parallel \( \newcommand\) • • • • • • • • • • Learning Objectives By the end of this section, you will be able to: • Explain how to determine the equivalent capacitance of capacitors in series and in parallel combinations • Compute the potential difference across the plates and the charge on the plates for a capacitor in a network and determine the net capacitance of a network of capacitors Several capacitors can be connected together to be used in a variety of applications. Multiple connections of capacitors behave as a single equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. Capacitors can be arranged in two simple and common types of connections, known as series and parallel, for which we can easily calculate the total capacitance. These two basic combinations, series and parallel, can also be used as part of more complex connections. The Series Combination of Capacitors Figure \(\PageIndex.\] This expression can be generalized to any number of capacitors in a series network. Example \(\PageIndex \nonumber\] Significance Note that in a series network of capacito...

The charge on 75 nF =1.65*10^- 5 C. _________________________ C) The total energy stored U =(1/2)CV^2=2.662*10^-3 J _________________________________ D) The energy stored in 35 nF capacitor.=8.47*10^-4 J The energy stored in 75 nF capacitor.=1.815*10^-3 J _________________________________ E) The potential difference across each capacitor is 220 V _________________________________ • I always remember these formulas which are for DC based on the resistor first. In series Rt = R1+R2+R3... In parallel 1/Rt = 1/R1+1/R2+1/R3 Since resistors and coils (inductors) have effect but pass electricity the formulas work the same. However capacitors block electricity, so the same formulas in reversed situations.

Homework Statement For the capacitor network shown in the Figure , the potential difference across ab is 12.0 V A) Find the total energy stored in this network. B) Find the energy stored in the 4.80μF capacitor. Homework Equations C = Q/V The Attempt at a Solution I'm honestly not sure where to start. I've tried playing around with the units but I haven't managed to find anything that gives me joules. I managed to get it I found that the equation for the energy stored in a capacitor is: U = 1/2*CV^2 I found C by breaking it up into the separate capacities. So, for the left most capacities I found their equilibrium capacitor, which I did by adding the reciprocal of the values: 1/8.60 + 1/4.80 = 0.324612 This becomes 3.080597 when you take it's receiprocal. Then I found the capacitor for the top-most capacitors. 1/6.20 + 1/11.8 = 0.24603 Taking the reciprocal gives me 4.06444 Then I added 4.06444 to 3.50 because these capacitors are parallel, whereas previously the capacitors had been in series. This gave me 7.56044 Then I did: 1/7.56044 + 1/3.080597 = 0.45687 For which I remember to take the receiprocal, thereby giving me 2.1887593 micro-F Plugging this into the formula gives me: U= 1/2*2.1887593 micro-F*12^2 U = 157.59 micro-J Thanks for your help The potential difference across that capacitor is not 12 V. You can look at the circuit with the two rightmost capacitor plus one that replaces all the left most ones. So you have a series circuit with three capacitors which shou...

students Hyderabad question potential difference across ab in the network shown in figure so here we have to find the potential difference across this point a and this point we now latest take the potential at point A is equals to be a OK so here and here and here the potential will be way because there is no resistance present between these points so there is no potential difference ok and hear the potential we have taken to be be be so there will be be be and here also it will be be be ok now we can see that the potential difference across this 30 ohm this 30 ohm and 10 30 ohm is equals to v a minus b b that means the three resistance R and parallel combination the equivalent for parallel combination is given by one upon our equivalent is equals to 1 upon 1 + 1 upon 2 + 1 upon 3 ok now the value of R1 R2 and R3 is equal here that is 30 ohm so latest put it here so we have when divided by 30 + 1 divided by 30 + 1 divided by 30 so it will be equals to 3 times of when divided by 30 that is 3/32 from this we have the value of our equivalent is equals to 33 and it is equals to 10 ohm ok that means they couldn't of this combination is Tenu ok now for finding the potential between A and B we have to find the current through the circuit latest draw the simple circuit here so in simple circuit we have the equivalent resistance between a and b is equals to 10 ohm that we have calculated before ok so it is point P is a point a and afterwards there is a resistance of 15 Ohm and here...

Hint: Use the formula for equivalent resistance of the two resistors connected in series and parallel arrangement. Also use the expression for Ohm’s law. First determine the equivalent resistance in the circuit using the formula for equivalent resistance in parallel and series arrangement. Then determine the current in the circuit using Ohm’s law. Finally, calculate the potential difference across AB using Ohm’s law. Formulae used: The equivalent resistance \[\] …… (2) The expression for Ohm’s law is given by \[V = IR\] …… (3) Here, \[V\] is the potential difference between the two ends of the conductor, \[I\] is the current in the conductor and \[R\] is the resistance of the conductor. Complete Step by Step Answer: We have asked to calculate the potential difference between the ends A and B given in the circuit diagram.Let us first redraw the given circuit diagram to simplify the given circuit. To calculate the potential difference between the ends A and B of the circuit, we need to calculate the net resistance and current in the circuit.Let us first determine the net resistance of three resistors connected in parallel.According to equation (1), we can write \[\dfrac\]. Hence, the correct option is C. Note:The students may only determine the equivalent resistance of the three resistors connected in parallel and then use Ohm’s law to determine the current and then potential difference across AB. But this is an incorrect way to solve the question. The students should calcul...