Tutor4Physics.com

 physics
homework help

math, chem help? click

 Ask questions, get answers

Home - Ask Questions - Latest Answers - Past Answers - Physics Terms - Physics Formulas - Tutorial - Exam tips
  Site Map - Free downloads - Home Work Helper - Physics Calculators - About Us - Testimonials
 Work Power Energy | 1D Motion | 2 & 3D Motion | Circular Motion | Laws of Motion | Gravitation | Electricity | Properties of Matter
In a region where there is an electric field, the electric forces do +8.0 x 10-19 J of work on an electron as it moves from point X to point Y. Which point, X or Y is at a higher potential? What is the potential difference, Vy - Vx , between point Y and point X?

W = qV

If W is positive, either both q and V are positive or both q and V are negative. Since electron ha a negative charge, V should be negative in this case (W is positive). Thus final potential must be less than the initial potential. Thus X is at a higher potential.

V = Vy - Vx

W = 8.0 x 10-19 J

q = -1.6 x 10-19 C

V = Vy - Vx = W/q = 8.0 x 10-19 / -1.6 x 10-19 = -5 Volt.

A parallel plate capacitor has a capacitance of 1.2nF. There is a charge of magnitude 0.800 µC on each plate. What is the potential difference between the plates? If the plate separation is doubled, while the charge is kept constant, what will happen to the potential difference?

C = 1.20 x 10-9 F

Q = 0.800 x 10-6 C

Q = CV

or V = Q/C = 0.800 x 10-6 / 1.20 x 10-9 = 0.667 x 10+3 V

The capacitance of a parallel plate capacitor

C = Aeo /d

where d is the plate seperation

If separation is doubled, C is halved.

If Q is kept constant, V = Q/C will be doubled.

  

In the circuit given below find the value of resistance R and then find the value of the maximum power to resistance R.


Q The ac generator in Fig below supplies 120 V (rms) at 60 Hz. With the switch open as in the diagram, the resulting current leads the generator emf by 20 degs. With the switch in position 1 the current lags the generator emf by 10 deg. When the switch is in position 2 the rms current is 2.0 A. find the values of R,L, and c.

Ev = 120V , f=60 Hz , w=2pif = 360 radians

Current I leads by 20 degs

Phase is negative, Xl < Xc

Tan (-20deg)= (Xl - Xc)/R = -0.36 ---- (1)

Or (Xc -Xl)/R =0.36

I lags by 10 degs and phase is positive

Xl > Xc1

Tan 10 degs = 0.18 = (Xl - Xc1)/R ------(2)

Z=Xc - Xl = Ev /Iv = 120/2 =60 ---------(3)

Substituting 3 in 1

60/R=0.36

R=60/0.36 = 500/3 ohms

Equation 2 implies wL-1/2wC = 0.06 x 500/3 = 30

Equation 3 implies 1/wC -wL=60 ----------(a)

wL = 30 + 1/2wC ------(b)

substituting in a

1/wC -30 - 1/2wC =60

1/2wC = 90 -------------(c)

C=0.15 µF

Substituting c in b

wL=30 + 90

L= 1/3 H

 

Q Find the magnetic field at the center of the loop shown.

B = (µo/4pi)(IL/r2)

B = 0

Since I1R1 = I2R2 or I1L1 = I2L2

Or B = (µo/4pir2)( I1L1 - I2L2) = 0

Q What is the back emf in a motor that draws 7.1 A from a 220 volt line if the field coils have a resistance of 200 ohms and are wound in parallel with an armature with a resistance of 3 ohms?

I=7.1 A , v = 220 V , R1 = 200 ohms, R2 = 3 ohms

Let E = back emf

I =(V-E)/R

1/R =1/3 +1/200 ie R = 600/203

E = V - IR = 220 - 21 = 199 V

Q. A horizontal rod 2 meters long is released in a region where there is a uniform horizontal magnetic induction of 0.6 Tesla perpendicular to the rod. How far has the rod fallen when the induced emf is 120V.

E=Blu

120 = 0.6 x 2x u

u = 100 m/s

initially u = o finally u =100 m/s

v2 - u2 =2as = 2gh

h= (1002 -0)/(2x10) =500 m