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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 22:30:16 IST
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hcv sum no 20 pg 199 plsssssssssss dudes solve it
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
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28 Mar 2008 22:40:38 IST
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ask a qustion
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28 Mar 2008 22:40:50 IST
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for fig in series..........
replace the 2 cells be a battery of voltage...12v.........and net resistance is 1+1+R......
u can get the current i1 =12/1+1+R.....
for fig in parallel.....
since the batteries r in parallel replace it by a net battery of emf 6 volt.......nd internal resistance .5 (as they r in parallel)..........now find current i2.......divide them nd u will get the answer.........
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28 Mar 2008 22:51:10 IST
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superb answer
could u also solve sum no 30
from the capacitor chapter could u solve sum no 47 and 25)d
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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28 Mar 2008 22:59:28 IST
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in the 30 ques if u observe closely......all the 3 resistances are in parallel hence the resistance is R/3............simplify the fig nd u wuld get it......
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28 Mar 2008 23:06:58 IST
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this is an easy sum but could u solve the capacitor sum
the question is
A CAPACITOR OF CAPACITANCE 5uF IS CHARGED TO A POTENTIOL DIFFRENCE 24V AND ANOTHER CAPACITOR OF CAPACITANCE IS CHARGED TO A POTENTIOL 12 V. THE POSITIVE PLATE OF THE FIRST CAPACITOR IS CONNECTED TO THE NEGATIVE PLATE OF THE SECOND CAPACITOR. FIND THE CHARGES APEEARING
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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28 Mar 2008 23:13:57 IST
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47 capacitor........
the potential across the capacitors should be same........
let q1,q2 be the charges.......
q1/5=q2/6
now since the polarity is reversed i.e + connected to -...........
q1+q2=120-72(where 120 nd 72 are the initial charges on the capacitor....)
solving the equations u wuld get the answer.........
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28 Mar 2008 23:18:25 IST
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regardin ques 25 d)
treat each branch seperately.....
charge from first branch ==6*4=24
charge frm 2nd branch=12*2=24
frm 3rd branch=24*1=24
now find the equivalent capacitance which is 4+2+1=7(connected in parallel).....
divide total charge=24+24+24=72
by the net capacitance=7
72/7=10.28
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28 Mar 2008 23:18:32 IST
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WHY WOULD THE POTENTIOLS BE SAME
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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28 Mar 2008 23:24:31 IST
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connecting wire is comman for both the capacitances hence potential is same...
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28 Mar 2008 23:26:30 IST
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COULD U SOLVE SUM NO 41B AND SUM NO 49 FROM THE RESISTOR SUM
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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28 Mar 2008 23:34:05 IST
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sum 41b
u can break the circuit as shown and then solve it by using series nd parallel method.........hope u understand the drawing.......
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28 Mar 2008 23:37:47 IST
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THE SUM NO 49 IS MORE IMPORTANT COULD U SOLVE IT FIRS
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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28 Mar 2008 23:43:46 IST
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let the resistance of the voltmeter be R..............
net resistance = 50r/(50+r) + 24........
current flowing===30/net resistance=I .......
so current through 50 and voltmeter resistance combination is also I...........
current through voltmeter=====50/(50+r)*I...........
multiply the current in voltmeter by its resistance r.....nd equate it to 18........solve it nd u will get the answer..........
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28 Mar 2008 23:44:45 IST
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HEY MAN ANUPUM UR A GENIUS IN ELECTROSTATICS
IF U COME TO GOIIT AT AFIXED TIME PLS LET ME KNOW IT
I
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
h |
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