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 ï¿½C 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
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)
Substituting c in b
wL=30 + 90
L= 1/3 H
Q Find the magnetic field at the center of the loop shown.
B = (ï¿½o/4pi)(IL/r2)
B = 0
Since I1R1 = I2R2 or I1L1 = I2L2
Or B = (ï¿½o/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
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.
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