Capacitors Questions and Answers

76184 For the current waveform shown below which is applied to a 2mF capacitor calculate the time at which the value of capacitor voltage Vc crosses 2000V Given that Vc 0 250V 3 marks 202 10 de A

2 Two capacitors are constructed Each is made of parallel plates that have an area of 1 75 x 10 4 m In capacitor 1 the plates are 0 040 m apart In capacitor 2 the plates are 0 020 m apart Both have Teflon between the plates Teflon has a dielectric constant of 2 1 What is the voltage on each capacitor if they are connected in series to a 12 0 V battery

Question 6 Find the equivalent capacitance for the below circuit S F 20 F 6 F 20 F 60 F

Potential difference of 100 0 V is applied across two capacitors in series as shown If C is a 6 F capacitor and has a voltage drop of 75 V across it what is the capacitance of C C2 100 V

A It takes 3 00J of energy to move a 1 50C charge from the positive side of a uniform electric field to the other What is the magnitude of the potential difference across plates B What is the magnitude of the electric field between those plates if they are 0 0350m apart

For the capacitor network shown in Figure 1 the potential difference across ab is 68 V Figure 150 p 120 p 1 of 1 Part F Find the energy stored in the 120 nF capacitor Express your answer with the appropriate units U Submit Part G V Submit HA Y Part H Value Request Answer Find the potential difference across the 150 nF capacitor Express your answer with the appropriate units HA Value Units Request Answer WC Units QWC

In the circuit below after a long time how much charge is on the 400 mF capacitor Select one O a 960 mC O b OC d 480 mc 600 mC 20 100mF 400mF 12V 500mF

6 27 Find the equivalent capacitance with respect to terminals a b for the circuits shown in Fig P6 27 Figure P6 27 a 20 nF F 40 V 8 nF 10 nF b 5 V 25 H HE 10 V 5 F HE 24 F 30 F J 48 nF HE 30 V 4 nF isminist 24 nF COM yk 30 nF s F 20 V 5 V He 36 V 10 V 12 F enver 18 F 2 VS

A parallel plate capacitor consists of two parallel square plates that have dimensions 1 0 cm by 1 0 cm If the plates are separated by 1 0 mm and the space between them is filled with teflon what is the capacitance of this capacitor The dielectric constant for teflon is 2 1 and 0 8 85 10 12 C2 N m2 0 89 pF O 1 9 pF O 0 44 pF O 2 1 pF

Figure 22 19 V 300 V a t 0 s b t 0 7 s 22 31 Assume that C in Fig 22 19 is completely discharged with S in Position 2 If S is moved to Position 1 how much is the capacitor voltage at the following time intervals c t 1 s d t 1 5 s e t 2 s f t 2 5 s R 1 MS 25 C 1 F

Figure 5 shows a system of capacitors, where the potential difference across ab is 38.0 V. (a) What is the equivalent capacitance of this system between a and b? (b) How much charge is stored by the system? (c) How much charge does the 18.0 nF capacitor store? (d) What is the potential difference across the 10.0 nF capacitor?

The rectangular-shaped current pulse shown in Fig. P6.21 is applied to a 0.1 μF capacitor. The initial voltage on the capacitor is a 15 V drop in the reference direction of the current. Derive the expression for the capacitor voltage for the time intervals in (a)-(d). a) 0 ≤ t ≤ 10 μs; b) 10 μs ≤ t ≤ 20 μs; c) 20 µs ≤ t ≤ 40 μs d) 40 us ≤ t <∞ e) Sketch v(t) over the interval -10 µs ≤ t ≤ 50 µs.

Three identical capacitors are connected in series across a potential source (battery). If a charge of Q flows into this combination of capacitors, how much charge does each capacitor carry? A) 3Q B) Q C) Q/3 D) Q/9

A large capacitance of 1.06 mF is needed for a certain application. (a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.98 um of Teflon, which has a dielectric constant of 2.1. (b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is 60 x 106 V/m (c) Find the maximum charge that can be stored. (d) Calculate the volume of Teflon alone in the capacitor.

Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of constant K is introduced between the plates. The energy which is stored becomes K times. Reason: The surface density of charge on the plate remains constant or unchanged.

You have an RC Circuit with a resistor and capacitor. You can't change the resistor, but you can change the capacitor. You need the circuit to take shorter (time) to charge to the maximum value. You will do this by changing the capacitor. In order to decrease the time, you should replace your capacitor with a new capacitor with a lower (smaller) capacitance.

Two square metallic plates of 1m side are kept 0.01 m apart, like a parallel plate capacitor, in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a speed of 0.001 m/s. Calculate the current drawn from the battery during the process.

You have an RC Circuit hooked up to a battery that you are testing for your company. This consists of a battery, resistor and capacitor all in series. You can assume the battery is ideal (no internal resistance). You need to know how long it takes (time) in order for the capacitor to store 84.9 % of the maximum possible energy for this circuit. The values for the battery, resistor and capacitor are below.

Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will : become zero increase decrease remains same

Capacitor A parallel plate capacitor has plates of area 2.5 × 10 m² and plate separation 1.2 × 104 m. Air fills the volume between the plates. What potential difference is required to establish a 4.0 µC charge on the plates?

Two capacitors connected in parallel having the capacities C₁ and C₂ are given 'q' charge, which is distributed among them. The ratio of the charge on C₁ and C2 will be: A. C₁ / C₂ B. C₂ / C₁ C. C₁C₂ D. 1/ C1C₂

You have 3 capacitors in series (there values are below) and they are hooked up to a battery of voltage = 6.00 V. They are allowed to reach electrostatic equilibrium. C1 = 64.0 x 10-6 F C2 = 15.0 x 10-6 F C3= 80.0 x 10-6 F What is the energy stored on C3? Give your answer in microJoules (J) and with an appropriate number of significant figures.

You have 3 capacitors in parallel (there values are below) and they are hooked up to a battery of voltage = 5.00 V. They are allowed to reach electrostatic equilibrium. C1 = 79.0 x 10-6 F C2 = 45.0 x 10-6 F C3 = 52.0 x 10-6 F What is the energy stored on C3? Give your answer in microJoules (J) and with an appropriate number of significant figures.

Consider a parallel-plate capacitor with plates of area A and with separation d. Part A Find F (V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor. Express your answer in terms of given quantities and €0. View Available Hint(s)

A parallel-plate capacitor has plates with area 2.30x10-2 m² separated by 1.10 mm of Teflon. Part A Calculate the charge on the plates when they are charged to a potential difference of 14.0 V. Express your answer in coulombs. VAE Submit Part B Request Answer ? Use Gauss's law to calculate the electric field inside the Teflon. Express your answer in newtons per coulomb. C

A 76.0 μF capacitor is initially uncharged. It is being charged through a 24.7 k resistor by a 3.00 V battery. a) What is the RC time constant? b) How long will it take for the capacitor to charge to 1/4 of its final charge? A.RC time constant is 1.88 s B.RC time constant is 3.08 s C.RC time constant is 1877 s D.RC time constant is 2.28 x 10-4 s E.RC time constant is none of these answers F. It will take 0.470 s G. It will take 0.540 s H. It will take 7.52 s I. It will take 6.53 s J. It will take 2.61 s

You need to design a capacitor capable of storing 4.0 x 10^-7 C of charge. At your disposal, you have a 100 V power supply and two metal plates, each of area 0.275 m². What is the limit of the separation of the plates?

In a circuit, a capacitor has potential difference AV, charge Q, and capacitance C. The potential difference is doubled. The capacitance: A. changes in ways impossible to predict with the given information. B. doubles. C. does not change. D. is divided in half.

The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by the distance between them. When a voltage of 84.4 V is put across the plates of such a capacitor an electric field strength of 4.0 kV/cm s measured. Write an equation that will let you calculate the distance d between the plates. Your equation should contain only symbols. Be sure you define each symbol.

A capacitor consists of two 7.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 160 V, then the battery is removed. 1)How much energy is stored in the capacitor? 2)How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?

Two 10 μF parallel plate capacitors are connected in parallel. (A) How much energy is stored if the voltage both capacitors is 100 V? (B) If the relative permittivity of one of the 10 μF parallel plate capacitors is 10. What is the ratio between the area of the plates and the distance that separates them?

A set of charged plates is separated by 8.08*10^-5 m. When 2.24*10^-9 C of charge is placed on the plates, it creates a potential difference of 855 V. What is the area of the plates? (The answer is *10^-5 m^2. Just fill in the number, not the power.)

A set of charged plates is separated by 2.22*10^-4 m. When 5.24*10^-9 C of charge is placed on the plates, it creates a potential difference of 240 V. What is the area of the plates? (The answer is *10^-4 m^2. Just fill in the number, not the power.)

An ideal parallel-plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected across a battery that maintains a constant potential difference between the plates. If the separation between the plates is now doubled, the magnitude of the charge on the plates will A. quadruple B. not change C. be cut in half D. be cut in fourth E. double

A 1.2 nF parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nF when the gap is filled by a dielectric. What is the dielectric constant of the dielectric? A. 2.5 B. 3.5 C. 5.5 D. 4.5

5 points Save Answe A parallel plate capacitor with a vacuum between the plates has a capacitance of 32.9 μF. When the space between the plates is filled with some material, the capacitance is 122 uF. What is the dielectric constant of the material?

The two plates of an empty (i.e. air-filled) parallel plate capacitor are connected to a battery. Sometime later, while the battery is still connected, a block of dielectric material (K=2 is inserted between the plates of the capacitor. What happens to the charge on the capacitor and the electric field between the plates? a) Charge remains constant, electric field remains constant b) Charge increases, electric field increases c) Charge increases, electric field remains constant d) Charge remains constant, electric field decreases e) Charge decreases, electric field decreases

Consider an isolated capacitor with a dielectric slab between its plates. The two plates have charge Q and -Q. Because the capacitor is not hooked to anything, the charge on the plates cannot change. What happens to the energy of the capacitor if you pull out the dielectric slab? A. It decreases. B. It does not change. C. It increases.

Suppose you have two identical capacitors, each with a capacitance C. You want to use them to store as much energy as possible. What is the best way to wire the capacitors together before connecting them to a potential difference? A. In parallel B. Not enough information given C. In series

Potential difference of 100.0 V is applied across two capacitors in series, as shown. If C₁ is a 6 µF capacitor and has a voltage drop of 75 V across it, what is the capacitance of C₂?

Five capacitors are connected to a 12V battery as shown, C₁=2µF, C₂=4µF, C3=1μF, C4=8μF, and C₁= 3µF i) Determine the equivalent capacitance between points A and B ii) Determine the charge on each capacitor.

To measure the capacitance C of a capacitor, you attach the capacitor to a battery and wait until it is fully charged. You then disconnect the capacitor from the battery and let it discharge through a resistor of resistance R. You measure the time T₁/2 that it takes the voltage across the resistor to decrease to half its initial value at the instant that the connection to the capacitor is first completed. You repeat this for several different resistors. You plot the data as T₁/2 versus R and find that they lie close to a straight line that has slope 4.00 uF. ▼ Part A What is the capacitance of the capacitor? Express your answer with the appropriate units. M

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. What is the potential difference between the plates?

For the capacitor network shown in (Figure 1), the potential difference across ab is 68 V. Find the total energy stored in the network. Find the energy stored in the 150 nF capacitor. Find the energy stored in the 120 nF capacitor.

A 5.00 pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 x 102 V. The electric field between the plates is to be no greater than 1.00 x 10¹ N/C. As a budding electrical engineer for Live-Wire Electronics, your task is to design the capacitor by finding what its physical dimensions and separation must be. What must the separation between the plates in the designed capacitor be? What must the radius of each plate in the designed capacitor be?

A switch that connects a battery to a 20μF capacitor is closed. Several seconds later you find that the capacitor plates are charged to +30μC. What is the emf of the battery?

A parallel-plate capacitor is formed from two 3.3 cm-diameter electrodes spaced 1.6 mm apart. The electric field strength inside the capacitor is 3.0 x 106 N/C. What is the magnitude of the charge (in nC) on each electrode?

We want to use the Earth's magnetic field to apply a torque on a current loop. Suppose the Earth's magnetic field at this location is 5*10^(-5) Tesla (0.5 Gauss) and we want to apply a maximum of 1.2 Newton-meters of torque (if the loop is oriented parallel to the magnetic field). The loop is circular with a radius of 1.5 meters. How much current do we need to apply to the loop? The loop is made of aluminum wire (circular in cross section) with a thickness of 0.02 meters (or a radius of 0.01 meters). Aluminum has a resistivity of 2.65*10^(-8) Ohm meters. How much power does it take to power the loop?

A charge Q = -6.51 mC is initially on the right plate, shown on the right, which has a voltage of VRight = +38.2 V. The charge moves to the left plate which has a voltage of VLeft = -38.2 V. What is the work done by the electric field in between the two plates moving the charge from the right- side plate to the left-side plate? Select one: a. -0.497 J b. -0.249 J c. +0.249 J d. +0.497 J

A 4500 resistor is connected across a charged 0.750 nF capacitor. The initial current through the resistor, just after the connection is made, is measured to be 0.249 A. What magnitude of charge was initially on each plate of this capacitor? How long after the connection is made will it take before the charge is reduced to of its maximum value?