We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>

Looking For Something at vustudents.ning.com? Click Here to Search

www.bit.ly/vucodes

+ Link For Assignments, GDBs & Online Quizzes Solution

www.bit.ly/papersvu

+ Link For Past Papers, Solved MCQs, Short Notes & More


Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion

How to Add New Discussion in Study Group ? Step By Step Guide Click Here.

Q # 1

 

A parallel plate air capacitor of capacitance 245pF has a charge of magnitude 0.148μC on each plate. The plates are 0.328mm apart.

(a) What is the potential difference between the plates?

(b) What is the area of each plate?

(c) What is the electric field magnitude between the plates?

(d) What is the surface charge density on each plate?                                           Marks 8

Q # 2

 

A cylindrical copper cable 1.50Km long is connected across a 220.0V potential difference.

(a) What should be its diameter so that it produces heat at a rate of 50.0W?

(b) What is the electric field inside the cable under these conditions?                    Marks 6

 

Q # 3

 

A person with body resistance between his hands of 10KΩ accidently grasps the terminal of a 14-KV power supply.

(a) If the internal resistance of the power supply is 2000Ω, what is the current through the person’s body?

(b) What is the power dissipated in his body?

(c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00mA or less?                                                                                                       Marks 6

 

<Good Luck>>>>>>>>>>>>>>

+ How to Follow the New Added Discussions at Your Mail Address?

+ How to Join Subject Study Groups & Get Helping Material?

+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?

+ VU Students Reserves The Right to Delete Your Profile, If?


See Your Saved Posts Timeline

Views: 1305

.

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

+ Click Here to Search (Looking For something at vustudents.ning.com?)

+ Click Here To Join (Our facebook study Group)

Replies to This Discussion

Please Discuss here about this assignment.Thanks

Attachments:

Please share your idea solution...

Waiting for the solution to come......................
plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz koi solution upload kar da plzzzzzzzzzzzz
Q3:
(a)The current 'I' that flows when a voltage 'V' is applied across a resistor 'R' is given by I = V/R.
Here, the total resistance is 10000 + 2000 = 12000 ohm.
The voltage applied across this resistance is 14000 V.
This makes a current equal to 14000/12000 = 7/6 amp flow through the person's body.
(b) The power P dissipated is given as P = V^2/R
We are only interested in the power dissipated through the person's body. This is given by (14000)^2/10000 = 19600 W
ec ka b third part nae

Q#2:

First, the value of the length of the cable will be converted to meters.

Therefore, the length of the cable is L = 1.5*10^3 m = 1500 m.

We know, from enunciation, the value of potential difference:V = 220 volts

The power is given and it's value is of  P = 50 Watts

P = V^2/R => R = V^2/P

Since the area of the  section of the cable is circular, w'ell recall the formula for the area of the circle:

A = pi*r^2 (1)

A = p*L/R (2), where p = 1.72/10^8 ohm/m

We'll equate (1) and (2) and we'll get:

pi*r^2 = p*L*P/V^2

r^2 = p*L*P/V^2*pi

r = sqrt(p*L*P/V^2*pi)

Diameter is d = 2*r = 2*sqrt(p*L*P/V^2*pi)

d = 2*sqrt1.72*10^-10*15*5*10^3/484

d = 2*sqrt 1.72*10^-7*75/22

d = 1.0325/100

d = 0.010325 meters

The diameter of the cable is of d = 0.010325 meters, such as it is producing heat at a rate of 50W.

 

and what is electric field ?? second part

Q#1:

(a)we have q=cv where q=0.148μC

(remember we only put the magnitude of

one plate and we don't double it) and

c=245pF, so

v(potential difference)=604.08 volt
(b)c=(8.85*10^-12)*A/d where d=0.328mm,

so A(area)=9.08*10^-3 .
(c)v=Ed so E=v/d,

E(electric field)=1.84*10^6 v/m
(d)surface charge density=q/A=1.63*10^-5

Muhammad_Mansoor_Waqas ap achnak achay lagnay lag gy ho
Thanks bro

PHY101 Assignment#04 Complete Solution Spring 2011

See the attached file please

Attachments:

RSS

Today Top Members 

© 2020   Created by +M.Tariq Malik.   Powered by

Promote Us  |  Report an Issue  |  Privacy Policy  |  Terms of Service

.