An Uncharged capacitor `C` is connected to a battery through a resistance `R`. Show that by the time the capacitor gets fully charged, the energy dissipated in ``R is the same as the energy stored in `C`.

Explanation

To solve the problem, we need to show that the energy dissipated in the resistor \( R \) is equal to the energy stored in the capacitor \( C \) when it is fully charged. Let's go through the steps systematically.

### Step 1: Understand the Circuit
We have a circuit consisting of a battery, a resistor \( R \), and an uncharged capacitor \( C \). When the capacitor is connected to the battery, it begins to charge through the resistor.

### Step 2: Energy Stored in the Capacitor
The energy stored in a capacitor when it is fully charged is given by the formula:
\[ U_C = \frac{1}{2} C V^2 \]
where \( V \) is the voltage of the battery.

### Step 3: Charge on the Capacitor
The charge \( Q \) on the capacitor when it is fully charged is given by:
\[ Q = C V \]

### Step 4: Total Energy Supplied by the Battery
The total energy supplied by the battery when the capacitor is fully charged is:
\[ U_B = Q \cdot V = (C V) \cdot V = C V^2 \]

### Step 5: Energy Dissipated in the Resistor
The energy dissipated in the resistor \( R \) can be found by subtracting the energy stored in the capacitor from the total energy supplied by the battery:
\[ U_R = U_B - U_C \]
Substituting the expressions we derived:
\[ U_R = C V^2 - \frac{1}{2} C V^2 = \frac{1}{2} C V^2 \]

### Step 6: Compare Energy Dissipated and Stored
From the previous steps, we have:
- Energy stored in the capacitor: \( U_C = \frac{1}{2} C V^2 \)
- Energy dissipated in the resistor: \( U_R = \frac{1}{2} C V^2 \)

Thus, we can conclude:
\[ U_R = U_C \]

### Conclusion
The energy dissipated in the resistor \( R \) is equal to the energy stored in the capacitor \( C \) when it is fully charged.

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