Question

In the circuit shown, the potential difference across the $3\mu F$ capacitor is V, and the equivalent capacitance between A and B is $C_{AB}$ :

• A. $C_{AB}=4\mu F$
• B. $C_{AB}=\frac{18}{11}\mu F$
• C. V=20 V
• D. V=40 V

Answer 1

Answer: (A), (D)

$C_{AB}=4\mu F$
V=40 V

$C_{AB}=2+\frac{3\times 6}{3+6}=2+2=4\mu F$
As 2 and (3,6) are in parallel so potential across 2 and (3,6) will be same i.e $60V$
$C_{eq}^{(3,6)}=\frac{3\times 6}{3+6}=2\mu F$ and $Q_{eq}^{3,6}=2\times 60=120\mu C$
As 3 and 6 are in series so charge on 3 and 6 is equal to equivalent charge $120\mu C$
The voltage across 3 is $V=\frac{120}{3}=40V$