A 5 μ F capacitor is connected in series with a 10 μ F capacitor. When a 300 V potential difference is applied across this combination, the energy stored in the capacitors is

Question

A 5 μ F capacitor is connected in series with a 10 μ F capacitor. When a 300 V potential difference is applied across this combination, the energy stored in the capacitors is (A) 15 J (B) 1.5 J (C) 0.15 J (D) 0.10 J

Tardigrade Admin · Accepted Answer

According to question the figure can be drawn as below

The equivalent capacitance,

\frac{1}{C _{eq}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

= \frac{1}{5} + \frac{1}{10}

= \frac{2 + 1}{10} = \frac{3}{10}

\Rightarrow C_{θq} = \frac{10}{3} μ F

Now, the energy stored in the capacitor is

U = \frac{1}{2} C V^{2}

= \frac{1}{2} \times \frac{10}{3} \times 1 0^{- 6} \times 300 \times 300

= \frac{3}{10 \times 2} = \frac{3}{20} = 0.15 J

Q. A 5μF capacitor is connected in series with a 10μF capacitor. When a 300V potential difference is applied across this combination, the energy stored in the capacitors is

15 J

1.5 J

0.15 J

0.10 J

According to question the figure can be drawn as below The equivalent capacitance, Ceq ​1​=C1​1​+C2​1​ =51​+101​ =102+1​=103​ ⇒Cθq​=310​μF Now, the energy stored in the capacitor is U=21​CV2 =21​×310​×10−6×300×300 =10×23​=203​=0.15J

Q. A $5 μ F$ capacitor is connected in series with a $10 μ F$ capacitor. When a $300 V$ potential difference is applied across this combination, the energy stored in the capacitors is