Ans.

(i)

Sol. Potential difference across C 4 is V 4 .

V 4 = V 1

∴ V 4 = 17.5 V

So the correct answer is

(ii)

Sol. Consider any branch, say, ZAY.

Charge on C 1 = C 1 V 1 =2 × 17.5 = 35µC

Since C 1 and C 2 are in series, charge on C 2 is also 35µC.

Hence the correct option in Q.3 is

Using Q = CV for C 2 ,

C 2 = C = = = 2.8 µF

Equivalent capacitance of branch XAY is the series equivalent of C 1 = 2µ and C 2 = C = 2.8 µF, i.e., = 1.17µF

Equivalent capacitance of branch XBY is the series equivalent of C 3 = C = 2.8µF and C 4 = 2µF i.e. 1.17µF. These branches are in parallel between X and Y. Hence, equivalent capacitance between X and Y = 1.17 + 1.17 = 2.34 µF

Hence the correct option in Q. 2 in

(iii)

Sol. Capacitor C 6 is connected in the part of circuit between X and A either in series or parallel with C 1 = 2µF. Let the equivalent capacitance of C 1 and C 5 i.e., the equivalent capacitance between X and A be C´ 1 .

Capacitor C6 is connected between A and Y. Obviously, the circuit then becomes a Wheatstone bridge. Further, since equivalent capacitance between X and Y is independent of the value of C 6 , it implies that the bridge is in the balanced condition and potentials at A and B are now equal, so that:

=

∴ = (C 2 = C 3 = 2.8µF, as or determined earlier)

C 1 = 3.92µF

C´ 1 is the equivalent capacitance of C 1 = 2µF and C 5 . Since C´ 1 > C 1 , therefore, we can conclude that C 1 and C 5 could not be connected in series. We know that series equivalent capacitance is less then each individual capacitance. Hence, C 1 and C 5 are are in parallel, so that

C´ 1 = C 1 + C 5

or C 5 = C´ 1 – C 1 = 3.92 – 2 = 1.92µF