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 , dʹ = 2 d ,
And area become half , Aʹ = A / 2 ,
So ,
Ø Cʹ = ( kAʹ£o ) / dʹ .
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 , Iʹ = 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 .
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