Let us now connect two more capacitors in the circuit. One of them, `C_(5)`, is connected in the part of the circuit between `X` and `A`. It connected between either in series or in parallel with `C_(1)`. The other, `C_(6)`, is connected between `A and B`. It is observed `X and Y` hass the the same malue `C_(5)` is.
Capacitance `C_(5)` is.

Options

A. `1.24 muF` in series with `C_(1)`
B. `1.92 muF` inh parallel with `C_(1)`
C. `2.28 muF` in series with `C_(1)`
D. `2.56 muF` in parallel with `C_(1)`

Answer

B

Explanation

Potential differnce across `C_(4)` is `V_(4)`. As discussed above
`V_(4)=V_(1)`
`V_(4)=17.5 V`
Consider any branch say, `XAY`
Since `C_(1)` and `C_(2)` are in series, charge on `C_(2)` is also `35 mu C`.
Using `Q = CV` for `C_(2)`
`C_(2) = C = (35)/(V_(2)) = (35)/(12.5) = 2.8 mu F`
Equivalent capacitance of branch `XAY` is the series equivalent of `C_(3)=C=2.8 muF` and `C_(4)=2 muF` i.e., `1.17 muF`. There branches are in parallel between `X` and `Y`, Hence equivalent capacitance between `X` and `Y` is `1.17+1.17=2.34 muF`.
Capacitor `C_(5)` is connected in the part of circuit between `X` and `A` either in series or parallel with `C_(1)=2 muF`. Let the equivalent capacitance of `C_(1)` and `C_(5)`, i.e., the equivalent capacitance between `X` and `A` be `C'_(1)`.
Capacitor `C_(6)` is connected between `A` and `B` 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 bridge is in the balanced condition and potentials at `A` and `B` are now equal, so that
`(C'_(1))/C_(3)=C_(2)/C_(4)` or `(C'_(1))/(2.8)=(2.8)/2` (`C_(2)=2.8 muF`, as or determined earlier)
or `C_(1)=3.92 muF`
`C'_(1)` is the equivalent capacitance of `C_(1)=2 muF` and `C_(5)`, Since `C'_(1)gtC_(1)`, we can conclude that `C_(1)` and `C_(5)` are in parallel, So each individual capacitance .Hence, `C_(1)` and `C_(5)` are in parallel, so
`C'_(1)=C_(1)+C_(5)` or `C_(5)=C'_(1)-C_(1)=3.92-2=1.92 muf`
`V'_(1)+V_(2)=30`
.
.
Since the bridge is in a balanced condition, `A` and `B` at same potential and `C_(6)` has zero charge. `C'_(1)` and `C_(2)` are in series. Therfore, charge on `C'_(1)` and `C_(2)` has the same value, so that potential differnce will be in the inverse ratio of capacitance
[`Q=CV` or `V=Q/C "hence for same" Q,Vprop1/C']`
`(V'_(1))/(V'_(2))=(2.8)/(3.92)~~0.71` or `V'_(1)=0.71 V_(2)`
From Eq. (iv), `0.71 V_(2)+V_(2)=30` or `V_(2)=17.5 V`
Hence `V'_(1)=30-V_(2)=12.5 V`
Since `C'_(1)`, is parallel equivalet of `C_(1)` and `C_(5)`, so potential differnce across each is `V'_(1)=12.5 V`.
Hence, charge on `C_(5)` is
`C_(5)V'_(1)=1.92 xx12.5=24 muC`.