KCET · Physics · Capacitance
When an additional charge of \(2 \mathrm{C}\) is given to a capacitor, energy stored in it is increased by \(21 \%\). The original charge of the capacitor is
- A \(30 \mathrm{C}\)
- B \(40 \mathrm{C}\)
- C \(10 \mathrm{C}\)
- D \(20 \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(20 \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Energy stored by a capacitor, \(U=\frac{1}{2} \frac{q^{2}}{C}\)
So, here \(U_{1}=\frac{1}{2} \frac{q_{1}^{2}}{C} ; \quad U_{2}=\frac{1}{2} \frac{q_{2}^{2}}{C}\)
\(\therefore \quad \frac{U_{1}}{U_{2}}=\left(\frac{q_{1}}{q_{2}}\right)^{2}=\left(\frac{q}{q+2}\right)^{2}\)
Given,
\(\frac{U_{2}-U_{1}}{U_{1}}=0.21\)
or
\(\frac{U_{2}}{U_{1}}=1.21 \Rightarrow\left(\frac{q}{q+2}\right)^{2}=1.21\)
On solving we, get
\(q=20 \mathrm{C}\)
So, here \(U_{1}=\frac{1}{2} \frac{q_{1}^{2}}{C} ; \quad U_{2}=\frac{1}{2} \frac{q_{2}^{2}}{C}\)
\(\therefore \quad \frac{U_{1}}{U_{2}}=\left(\frac{q_{1}}{q_{2}}\right)^{2}=\left(\frac{q}{q+2}\right)^{2}\)
Given,
\(\frac{U_{2}-U_{1}}{U_{1}}=0.21\)
or
\(\frac{U_{2}}{U_{1}}=1.21 \Rightarrow\left(\frac{q}{q+2}\right)^{2}=1.21\)
On solving we, get
\(q=20 \mathrm{C}\)
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