Q.
In the following diagram, the charge and potential difference across 8 $\mu F$ capacitance will be respectively
A) $320\ \mu\ C,\ 40\ V$
B) $420\ \mu\ C,\ 50\ V$
C) $214\ \mu\ C,\ 27\ V$
D) $360\ \mu\ C,\ 45\ V$
Explanation
(c) Given circuit can be redrawn as follows capacitors, $9\ \mu F,\ 9\mu F$ and $7\mu F$ are short circuited. So they are deleted
[image:1_image265.emf]
$V_{1} + V_{2} = 40V$
and $\frac{V_{1}}{V_{2}} = \frac{36}{18} = 2$
Hence $V_{1} = \frac{80}{3}V$
and $V_{2} = \frac{40}{3}V$
charge on $8\mu F$ capacitor $= 8 \times \frac{80}{3} = 213.3\mu F \approx 214\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.