MHT CET · Physics · Capacitance
If the charge on the capacitor is increased by 2 coulombs, the energy stored in it increase by \(21 \%\). the original charge on the capacitor is
- A 10 C
- B 15 C
- C 20 C
- D 5 C
Answer & Solution
Correct Answer
(C) 20 C
Step-by-step Solution
Detailed explanation
Concept: energy stored in capacitor is given by
\(U=\frac{q^2}{2 c}\)
Where, \(q\) is charge and \(c\) is capacitance.
Now, if change is increased by \(\delta\), the energy stored is
\(\Rightarrow U_n=\frac{(q+\delta)^2}{2 c}\)
Given, \(\delta=2 \mathrm{C}, U_n=1.21 \mathrm{U}\)
\(\begin{aligned}
& \therefore 1.21\left(\frac{q^2}{2 c}\right)=\frac{(q+\delta)^2}{2 c} \\
& \Rightarrow\left(\frac{q+\delta}{q^2}\right)^2=1.21 \\
& \Rightarrow\left(\frac{q+\delta}{q}\right)=1.1 \\
& \Rightarrow \frac{\delta}{q}=0.1
\end{aligned}\)
\(\Rightarrow q=\frac{\delta}{0.1}=20 \mathrm{C}\)
\(U=\frac{q^2}{2 c}\)
Where, \(q\) is charge and \(c\) is capacitance.
Now, if change is increased by \(\delta\), the energy stored is
\(\Rightarrow U_n=\frac{(q+\delta)^2}{2 c}\)
Given, \(\delta=2 \mathrm{C}, U_n=1.21 \mathrm{U}\)
\(\begin{aligned}
& \therefore 1.21\left(\frac{q^2}{2 c}\right)=\frac{(q+\delta)^2}{2 c} \\
& \Rightarrow\left(\frac{q+\delta}{q^2}\right)^2=1.21 \\
& \Rightarrow\left(\frac{q+\delta}{q}\right)=1.1 \\
& \Rightarrow \frac{\delta}{q}=0.1
\end{aligned}\)
\(\Rightarrow q=\frac{\delta}{0.1}=20 \mathrm{C}\)
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