Question

Two capacitors of 3 uF and 5 uF are in series across a 10 V source. Find the charge on these
capacitors and the potential difference across each. (ANS: Charge on the capacitors,
Q = 18.75x10-6 C, Potential difference across each, V1 = 6.25 V, V2 = 3.75 V)​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

Given :-

$\sf{C1 = 3 \: uF}$

$\sf{C2 = 5 \: uF}$

$\sf{V = 10 \: volt}$

Solution :-

Firstly ,

$\sf{When \: , 3 \: uF \: and \: 5 \: uF \: are \: in \: series-}$

$\sf{Total \: capicity-}$

$\sf{\dfrac{1}{Cs} =\dfrac{1}{C1} + \dfrac{1}{C2}}$

$\sf{Cs =\dfrac{C1\times C2}{C1 + C2}}$

$\sf{Cs =\dfrac{3\times 5}{(3 + 5)}}$

$\sf{Cs =\dfrac{15}{8} \: uF}$

We know that -

$\sf{Cs = \dfrac{Q}{V}}$

$\sf{Q = Cs\times V}$

$\sf{Q = \dfrac{15}{8}\times 10}$

$\sf{Q =\dfrac{15}{4}\times 5}$

$\sf{Q =\dfrac{75}{4}}$

$\sf{Q = 18.75 \: uC}$

$\sf{\sf\boxed{Q = 18.75\times 10^{-6} \: C}}$

Now ,

$\sf{Potential \: difference \: through \: 3 \: uF -}$

$\sf{V1 =\dfrac{Q}{C1} =\dfrac{18.75 \: uC}{3 \: uF}= 6.25 \: volt}$

Then ,

$\sf{Potential \: difference \: through \: 5 \: uF -}$

$\sf{V2 =\dfrac{Q}{C2} =\dfrac{18.75 \: uC}{5 \: uF}= 3.75 \: volt}$