Energies stored in capacitor and dissipated during charging a capacitor bear a ratio

Options

A. `1:1`
B. `1:2`
C. `1:1//2`
D. `2:1`

Explanation

To solve the problem of finding the ratio of energies stored in a capacitor and the energy dissipated during the charging of the capacitor, we can follow these steps:

### Step 1: Understand the Energy Supplied by the Battery
When a capacitor of capacitance \( C \) is charged by a battery with an electromotive force (emf) \( V \), the energy supplied by the battery can be calculated using the formula:
\[ E_B = Q \cdot V \]
where \( Q \) is the total charge transferred to the capacitor.

### Step 2: Relate Charge to Capacitance
The total charge \( Q \) on the capacitor can be expressed in terms of capacitance and voltage:
\[ Q = C \cdot V \]
Substituting this into the energy supplied by the battery gives:
\[ E_B = (C \cdot V) \cdot V = C \cdot V^2 \]

### Step 3: Calculate the Energy Stored in the Capacitor
The energy stored in the capacitor \( E_C \) can be derived from the formula:
\[ E_C = \frac{1}{2} C V^2 \]
This formula arises from integrating the work done to charge the capacitor.

### Step 4: Determine the Energy Dissipated
The energy dissipated during the charging process \( E_H \) can be found by subtracting the energy stored in the capacitor from the energy supplied by the battery:
\[ E_H = E_B - E_C \]
Substituting the expressions we found:
\[ E_H = C V^2 - \frac{1}{2} C V^2 = \frac{1}{2} C V^2 \]

### Step 5: Find the Ratio of Energies
Now we can find the ratio of the energy stored in the capacitor \( E_C \) to the energy dissipated \( E_H \):
\[ \text{Ratio} = \frac{E_C}{E_H} = \frac{\frac{1}{2} C V^2}{\frac{1}{2} C V^2} = 1 \]
Thus, the ratio of the energy stored in the capacitor to the energy dissipated during charging is:
\[ \text{Ratio} = 1:1 \]

### Final Answer
The ratio of the energy stored in the capacitor to the energy dissipated during charging is \( 1:1 \).

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