Question

A capacitor of 6uF charged to 40 volt connected in parallel across an
uncharged capacitor of capacitance of 6uF capacitor then the common
potential is​

Answer 1

Answer:

.3×10ki power -6

Explanation:

see thi photo

Answer 2

Answer:

The common potential of capacitors is 20V

Explanation:

Given that,

Capacitance of charged capacitor=6uF

Capacitance of uncharged capacitor=6uF

Voltage of capacitor=40V

The charge on the first capacitor is given as

$q=CV$

Substituting known values,we get

$q=6 \times 10 {}^{ - 6} \times 40 = 24 \times 10 {}^{ - 5}C$

The common potential can be found using relation

$V _{c} = \frac{total \: charge}{total \: capacitance}$

Therefore,substituting known values we get

$V _{c} = \frac{24 \times 10 {}^{ - 5} }{(6 + 6) \times 10 {}^{ - 6} } = \frac{24 \times 10 {}^{ - 5} }{12 \times 10 {}^{ - 6} } = 20V$

Therefore,the common potential of capacitors is 20V

#SPJ3