Problem 134 - A Complete Resource Book in Physics for JEE Main

Question

Assertion: When an uncharged capacitor of capacitance \(C\) is charged by a cell of emf \(V\), the energy stored by capacitor is \(\frac{1}{2} C V^{2}\), and energy supplied by battery is \(C V^{2}\). Reason: In charging an uncharged capacitor, energy is lost in the form of heat. (A) A (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) \(\mathrm{D}\)

Step-by-Step Solution

Answer

The correct answer is (A). The energy stored in the capacitor is equal to \(\frac{1}{2}CV^2\) and the energy supplied by the battery is equal to \(CV^2\). There is energy loss as heat during the charging process, which is supported by the efficiency being less than 1.

Step 1: Find the energy stored in the capacitor after charging

The energy stored in a charged capacitor is given by the formula: \(E_{stored} = \frac{1}{2} C V^2\) Where: \(E_{stored}\) = energy stored in the capacitor, \(C\) = capacitance of the capacitor, \(V\) = voltage across the capacitor. In our case, we have an uncharged capacitor with capacitance C, charged by a cell with emf V.

Step 2: Calculate the energy supplied by the battery

When the capacitor is being charged, the energy supplied by the battery is given by the formula: \(E_{supplied} = Q V\) Where, \(E_{supplied}\) = energy supplied by the battery, \(Q\) = charge on the capacitor after charging, and \(V\) = emf of the battery. Since \(Q = CV\), the equation becomes: \(E_{supplied} = C V^2\)

Step 3: Determine the energy lost as heat

If some energy is lost as heat while charging the capacitor, the efficiency can be represented as: Efficiency = \( \frac{E_{stored}}{E_{supplied}}\) Given the formulas for \(E_{stored}\) and \(E_{supplied}\), the efficiency is: Efficiency = \(\frac{\frac{1}{2}CV^2}{CV^2} = \frac{1}{2}\) Since the efficiency is less than 1, this means that there is energy loss as heat during the charging process. Now, by comparing our results to the given choices: (A) A - The energy stored in the capacitor is equal to \(\frac{1}{2}CV^2\) and the energy supplied by the battery is equal to \(CV^2\) which matches with our analysis. (B) B - Does not provide an answer. (C) C - Does not provide an answer. (D) D - Does not provide an answer. The correct answer is (A).

Key Concepts

Energy Stored in Capacitors

Capacitors are essential elements in electronic circuits, storing energy in an electric field when connected to a power source. When a capacitor with capacitance \( C \) is connected to a voltage source \( V \), it begins to store energy. The energy stored in a fully charged capacitor is given by the formula:

• \( E_{stored} = \frac{1}{2} C V^2 \)

This equation tells us how much energy is held within the capacitor's electric field. It's derived from the work done in transferring charge to the capacitor. Here:

• \( C \) is the capacitance, which defines the capacity of a capacitor to hold charge.
• \( V \) is the voltage or potential difference applied across the capacitor's plates.

The energy is stored as electric potential energy, and it's vital for applications like powering a flash in a camera or maintaining power supply in circuits during brief outages.
Understanding this concept helps in designing circuits where capacitors need to hold a specific amount of energy for various applications.

Charging Capacitors

Charging a capacitor entails moving electrons from one plate of the capacitor to the other, which builds up charge and voltage across the plates. When a battery (or another emf source) is connected to a capacitor, current flows until the capacitor is charged to the battery's voltage.

• During this process, the charge \( Q \) on the capacitor becomes equal to \( C \times V \).
• The energy supplied by the battery to charge the capacitor can be calculated using the formula:
  \( E_{supplied} = C V^2 \)

However, not all the energy supplied by the power source is stored in the capacitor due to resistive elements in the circuit, which dissipate some energy as heat.
This charging process and energy transfer are crucial in understanding the efficiency and performance of capacitors in electronic circuits. It involves concepts of potential difference and charge movement, foundational to the operation of most electronic devices.

Energy Efficiency in Capacitors

When charging capacitors, it's important to consider how efficiently the supplied energy is converted into stored energy within the capacitor itself. Energy efficiency in capacitors can be assessed by comparing the energy stored to the energy supplied.The efficiency is given by the ratio:

• Efficiency = \( \frac{E_{stored}}{E_{supplied}} \)

In the case of capacitors, this typically results in:

• Efficiency = \( \frac{\frac{1}{2}CV^2}{CV^2} = \frac{1}{2} \)

This implies that only half of the supplied energy gets stored in the capacitor, with the other half often lost as heat due to resistive elements during charging.
Understanding energy efficiency helps in designing more effective circuits and in enhancing power management in electronics. It is vital for applications where energy conservation is essential, as it directly impacts the performance and longevity of electronic devices.