MHT CET · Physics · Capacitance
An uncharged capacitor is connected to a battery. While charging the capacitor, how much is the energy lost, from the energy supplied by the battery?
- A \(50 \%\)
- B \(75 \%\)
- C \(100 \%\)
- D \(25 \%\)
Answer & Solution
Correct Answer
(D) \(25 \%\)
Step-by-step Solution
Detailed explanation
Consider the following diagram:
\(Q=C V \)
\( E_{\text {capocitor }}=\frac{Q^2}{2 C}=\frac{1}{2} C V^2 \)
\( W=\text {Work done by battery }=Q V=C V^2 \)
\( \therefore E_{\text {loss }}=(W)-\left(E_{\text {capocitor }}\right)=C V^2-\frac{1}{2} C V^2\) \(=\frac{1}{2} C V^2=\frac{W}{2}\)
This is \(50 \%\) off the energy supplied by the battery
\(Q=C V \)
\( E_{\text {capocitor }}=\frac{Q^2}{2 C}=\frac{1}{2} C V^2 \)
\( W=\text {Work done by battery }=Q V=C V^2 \)
\( \therefore E_{\text {loss }}=(W)-\left(E_{\text {capocitor }}\right)=C V^2-\frac{1}{2} C V^2\) \(=\frac{1}{2} C V^2=\frac{W}{2}\)
This is \(50 \%\) off the energy supplied by the battery
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