Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 18

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

A
C

B
nC

C
(n-1)C

D
(n+1)C

Solution

The given arrangement becomes an arrangement of (n-1) capacitors connected in parallel. So C_R = (n-1)C.
Question 2
1 / -0

To form a composite 16μF, 1000V capacitor from a supply of identical capacitors marked 8μF, 250V, we require a minimum number of capacitors

A
40

B
32

C
8

D
2

Solution

Suppose C = 8μF, C′ = 16μF′ and V = 250 V, V' = 1000V

Suppose m rows of given capacitors are connected in parallel and each row contains n capacitors then potential difference across each capacitor V = V′/n and equivalent capacitance of network C′ = mC.n. On putting the values we get n = 4 and m = 8. Total capacitors = n x m = 4 x 8 = 32