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Find the equivalent capacitance across $A$ $\&$ $B$ ........$\mu f$
$\frac{{28}}{3}$
$7.5$
$15$
none
Solution
The Capacitance $23 \mu f$ and $12 \mu f, 1 \mu f$ and $1 \mu f$ are connected in paralled so their equivalent capacitance is : $(23+12) \mu f,(1+1) \mu f$ respectively i,e
$35 \mu f$ and $2 \mu f$
Now the whole circuit is in form of wheatsfftone bridge with:
$C_{1}=35 \mu f, C_{2}=\mu f, C_{3}=10 \mu f$ and $C_{4}=2 \mu f$
Clearly, $\frac{C_{1}}{C_{2}}=\frac{C_{3}}{C_{4}}=5$
So $13 \mu f$ capacitor is neglected Now equivalent capacitance of $C_{1}$ and $C_{2}$ in series $=\frac{C_{1} C_{2}}{C_{1}+C_{2}} \mu f$ Equivalent capacitance of $C_{3}$ and $C_{4}$ in series $=\frac{10 \times 2}{10+2}=\frac{20}{12} \mu f$ Now these two equivalent capacitances are connected in parallel so their equivalent is:
$C_{e q}=\frac{245}{42}+\frac{20}{12}=\frac{15}{2} \mu f$
