A series ac circuit contains a 290 resistor, a 15.0 mH

A series ac circuit contains a 290 resistor, a 15.0 mH inductor, a 4.10 F capacitor, and an ac power source of voltage amplitude 45.0 Voperating at an angular frequency of 360 rad/s .

A) What is the power factor of this circuit?

B) Find the average power delivered to the entire circuit.

C) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

Solution

a) z = sqrt(R^2+(XC-XL)^2)

XL = 360*15*10^-3 = 5.4 ohm

XC = 1/WC = 677.5 ohm

Z = 731.9 ohm

power factor = R/Z = 0.396

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An air-filled parallel-plate capacitor has an area of 1

An air-filled parallel-plate capacitor has an area of 1.8 m2 and a capacitance of 1.4 nF.

Part A

What is the distance between the plates?

Part B

If the capacitor is connected to a 34 V battery, what is the electric field inside the plates?

Part C

If the capacitor is connected to a 34 V battery, what is the magnitude of the charge on each plate?

Part D

When the capacitor is connected to a 34 V battery, what is the energy stored in the capacitor?

Part E

The battery is disconnected from the capacitor so that the charge stays on each plate. You push the plates inwards so that the distance between the plates is half of what it was. What is the energy stored on the plates now?

Part F

How much work (positive or negative) are you doing on the plates moving them inwards? Why does this work make sense in terms of the force and displacement you apply?

Solution

area A=1.8 m^2

capacitance, C=1.4 nF

A)

C=eo*A/d

1.4*10^-9=8.89*10^-12*1.8/d

===> d=1.14*10^-2 m or d=1.14cm

distance between the plates, d=1.14cm

B)

V=34 V

use,

E=V/d

E=34/(1.14*10^-2)

E=2982.46 V/m

C)

q=C*V

q=1.4*10^-9*34

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You hike two thirds of the way to the top of a hill at

You hike two thirds of the way to the top of a hill at a speed of 3.0 and V run the final third at a speed of 6 omit/hr what way your average speed?

Solution

Time taken from A to B = D / S1

Time taken from B to A = D / S2

Total distance travelled = 2D

Therefore, an average speed is given by,

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Consider a square which is 1.0 m on a side. Charges are

Consider a square which is 1.0 m on a side. Charges are placed at the corners (marked A,B,C D) of the square as shown in the figure.(Figure 1) Q = 10 C and q = 2 C The distance of each corner to the center of the square is 0.707 m.

Part A

What is the magnitude of the electric field at the center of the square?

Express your answer using three significant figures

Part B

What is the direction of the electric field at the center of the square. (angle measured in degrees CCW from the positive x axis in the usual sense)

Part C

What is the electric potential at the center of the square?

Part D

If a 1C charge is placed at the center of the square, what is the x-component of the electrostatic force on the charge?

Part E

If a 1C charge is placed at the center of the square, what is the y-component of the electrostatic force on the charge?

Part F

If a 1C charge is placed at the center of the square, how much work (positive or negative) would you have to do to remove it to infinity?

(THANK YOU FOR TAKING THE TIME TO ANSWER THIS!)

.

+0 d=1 m -q D d=1 m a = 0.71 m +q QA

Solution

a)

r = distance of each charge from centre = 0.707 m

At the center electric field due to charge Q placed at A and C cancels out being equal and opposite in direction

the electric field by charges at B and D is in same direction and equal in magnitude

hence net electric field at centre is given as

E = k q/r2 + k q/r2 = 2 (9 x 109) (2 x 10-6)/(0.707)2 = 7.2 x 104 N/C

b)

direction : electric field is towards the charge located at D

angle = 90 + 45 = 135 CCW

c)

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