Question

Two capacitors of 3uF and 6uF are
connected in series across a potential
difference of 120 Volt. Then the
potential difference across 3uF
capaciter is-​

Answer 1

Answer:

Here is your answer: 80 volt

Explanation:

Qnet = Cnet x Volt

or, Q = CV

Cnet = CxC' / C+C'

= 6x3 / 6+3

= 18 / 9

= 2 micro Faraday

then, Qnet = Cnet x Volt

= (2 x 10^-6 ) x (120v)

= 240 micro coulomb

then, Qnet = C' V'

V' = Qnet / C'

= 240 / 3

V' = 80 volt Ans.

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