KCET · Physics · Current Electricity
A system of \( 2 \) capacitors of capacitance \( 2 \mu \mathrm{F} \) and \( 4 \mu \mathrm{F} \) is connected in series across a potential difference of \( 6 \mathrm{~V} \). The electric charge and energy stored in the system are
- A \( 10 \mu \mathrm{C} \) and \( 30 \mu \mathrm{J} \)
- B \( 36 \mu \mathrm{C} \) and \( 108 \mu \mathrm{J} \)
- C \( 8 \mu \mathrm{C} \) and \( 24 \mu \mathrm{J} \)
- D \( 1 \mu C \) and \( 3 \mu J \)
Answer & Solution
Correct Answer
(C) \( 8 \mu \mathrm{C} \) and \( 24 \mu \mathrm{J} \)
Step-by-step Solution
Detailed explanation
The effective capacitance of system of two given capacitors is
\(=\frac{1}{2 \mu F}+\frac{1}{4 \mu F}=\frac{(2+1)}{4 \mu F}=\frac{3}{4} \)
\(\Rightarrow C=\frac{4}{3} \mu F\)
Now, using \( Q=C V \)
\(Q=\frac{4}{3} \times 10^{-6} \mathrm{~F} \times 6 \mathrm{~V}=8 \times 10^{-6} \mathrm{C}\)
Therefore, electric charge \( =8 \mu \mathrm{C} \)
Now, energy stored \( =\frac{1}{2} C V^{2}=\frac{1}{2} \times \frac{4}{3} \times 10^{-6} \times 6 \times 6 \)
\( =2 \times 10^{-6} \times 2 \times 6=24 \times 10^{-6} \mathrm{~J} \)
Therefore, energy stored \( =24 \mu \mathrm{J} \)
\(=\frac{1}{2 \mu F}+\frac{1}{4 \mu F}=\frac{(2+1)}{4 \mu F}=\frac{3}{4} \)
\(\Rightarrow C=\frac{4}{3} \mu F\)
Now, using \( Q=C V \)
\(Q=\frac{4}{3} \times 10^{-6} \mathrm{~F} \times 6 \mathrm{~V}=8 \times 10^{-6} \mathrm{C}\)
Therefore, electric charge \( =8 \mu \mathrm{C} \)
Now, energy stored \( =\frac{1}{2} C V^{2}=\frac{1}{2} \times \frac{4}{3} \times 10^{-6} \times 6 \times 6 \)
\( =2 \times 10^{-6} \times 2 \times 6=24 \times 10^{-6} \mathrm{~J} \)
Therefore, energy stored \( =24 \mu \mathrm{J} \)
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