1
) what will be the equivalent
capacitance ?
A ) 1.5 C , B
) 1 C ,
C )
2 C , D ) None of these ,
Correct
Answer : - B ) 1 C ,
Solution
: -
Formulas
: -
Capacitance
in series : - 1/Cs = 1/C1
+ 1/C2 + 1/C3 + ……………… ,
Capacitance
in parallel : - Cp = C1 + C2
+ C3 + ……………… ,
So ,
Cp = 2C + 2C = 4C ,
So , we have
So ,
1/Ce = 1/4C + 1/2C + 1/4C ,
Therefore
, Ce = 1C , Answer .
2
) The capacitance of parallel plate
capacitor is ................ ( C = (kA£o)/d
) .
3
) In parallel plate capacitor , what would be new capacitance if separation (d)
is double and area in halved ?
A ) Becomes double , B
) Remain same ,
C ) Become one – forth , D ) None of these ,
Correct
Answer : - C ) Become one – forth ,
Solution
: -
Formula
: -
For
parallel plate capacitor , capacitance is :-
C
= ( kA£o ) / d .
If separation between slits become double , dÍ´ =
2 d ,
And area become half , AÍ´ = A / 2 ,
So ,
Ø Cʹ = (
kAÍ´£o ) / dÍ´ .
Ø Cʹ = k(A
/ 2)£o / 2d .
Ø Cʹ = (
kA£o ) / 4d . , So ,
Ø Cʹ = C
/ 4 , Answer .
4
) If current is increased by 100 times.
What would be effect on energy stored in inductor?
A ) increase by 1000 times , B
) increase by 10000 times ,
C ) No change , D
) None of these ,
Correct
Answer : - B ) increase by 10000 times
,
Solution
: -
Formula
: -
Energy
stored in inductor : -
E = (1/2)L(I)^2 ,
·
I
= current in inductor ,
·
L
= self inductance ,
So , if current is increased by 100 times , IÍ´ = 100 I
,
So energy
stored in inductor will be : -
Ø Eʹ = (1/2)L(Iʹ)^2 ,
Ø Eʹ = ( 1
/ 2 ) * L * ( 100 I )^2 ,
Ø Eʹ = ( 1 / 2 ) * L * ( 10000 I^2 ) = (10000)(1/2)LI^2 ,
Ø Eʹ = ( 10000 ) * E . Answer .
SO
, it will increased by ten thousand times .
Now go to online quiz section and check your preparation by this link : -
