Question

Two capacitors of capacitance 10 uF and 20 uF are connected in series
with a 6V battery. If E is the energy stored in 20 uF capacitor what will be
the total energy supplied by the battery in terms of E.​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{ANSWER :-}}}}} \\ \\ \bigodot{ \blue{Given :}} \\ \\ \implies \: two \: capacitors \: of \: 10 \: uF \: and \: 20 \: uF \: are \: </p><p>connected \: in \: series \: with \: 6v \: battery \\ \implies \: energy \: stored \: in \: 20 \: uF \: capacitor \: is \: E \\ \\ \bigodot{ \blue{To \: Find :}} \\ \\ \implies \: total \: energy \: supplied \: by \: battery \\ \\ \bigodot \blue{Formula :} \\ \\ \implies \: effective \: series \: capacitance \: is \: given \: by \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{cs = \frac{c1 \times c2}{c1 + c2}}} \\ \\ \implies \: energy \: supplied \: by \: battery \: \boxed{ \green{u = \frac{1}{2} cs ×{v}^{2}}} \\ \\ \implies \: energy \: stored \: in \: capacitor \: \boxed{ \blue{u = \frac{ {q}^{2} }{2c}}} \\ \\ \bigodot \: \blue{Calculation :} \\ \\ \implies \: cs = \frac{10 \times 20}{10 + 20} = \frac{20}{3} uF \\ \\ \implies \: charge \: on \: capacitor = cs \times v = \frac{20}{3} \times 6 = 40 \: uC \\ \\ \implies \: energy \: stored \: in \: 20 \: uF \: capacitor = \frac{ {q}^{2} }{2c} = \frac{ {40}^{2} }{2 \times 20} = 40 \: uJ \: = E \\ \\ \implies \: energy \: supplied \: by \: battery \: = \frac{1}{2} ×\frac{20}{3} ×{6}^{2} = 120 \: uJ = 3E \\ \\ \\ \implies \: so \: energy \: supplied \: by \: battery \: is \: \boxed{ \boxed{ \red{u = 3E = 120 \: uJ}}}$