A condenser of capacity `C` is charged to a potential difference of `V_(1)`. The plates of the condenser are then connected to an ideal inductor of inductance `L`. The current through the inductor when the potential difference across the condenser reduces to `V_(2)` is

Options

A. `((C(V_(1)-V_(2))^(2))/(L))^(1/2)`
B. `(C(V_(1)^(2)-V_(2)^(2)))/(L)`
C. `(C(V_(1)^(2)+V_(2)^(2)))/(L)`
D. `((C(V_(1)^(2)-V_(2)^(2)))/(L))^(1/2)`

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Initial Conditions
A capacitor with capacitance \( C \) is charged to a potential difference \( V_1 \). The initial charge \( Q_0 \) on the capacitor can be calculated using the formula:
\[ Q_0 = C \cdot V_1 \]

### Step 2: Connect the Capacitor to the Inductor
When the capacitor is connected to an ideal inductor of inductance \( L \), it begins to discharge, creating an oscillatory current in the circuit.

### Step 3: Define the Charge at Any Time
The charge \( Q \) on the capacitor at any time \( t \) during the discharge can be expressed as:
\[ Q = Q_0 \cos(\omega t) \]
where \( \omega \) is the angular frequency of the oscillation, given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 4: Relate the Charge to the Potential Difference
Initially, the potential difference across the capacitor is \( V_1 \), and it reduces to \( V_2 \). The relationship between charge and potential difference is given by:
\[ Q = C \cdot V \]
Thus, when the potential difference is \( V_2 \), we have:
\[ Q = C \cdot V_2 \]

### Step 5: Set Up the Equation
From the expression for \( Q \):
\[ C \cdot V_2 = Q_0 \cos(\omega t) \]
Substituting \( Q_0 = C \cdot V_1 \):
\[ C \cdot V_2 = C \cdot V_1 \cos(\omega t) \]
Dividing both sides by \( C \) (assuming \( C \neq 0 \)):
\[ V_2 = V_1 \cos(\omega t) \]
Thus, we can express \( \cos(\omega t) \) as:
\[ \cos(\omega t) = \frac{V_2}{V_1} \]

### Step 6: Find the Current Through the Inductor
The current \( I \) through the inductor is given by:
\[ I = -\frac{dQ}{dt} \]
Calculating \( \frac{dQ}{dt} \):
\[ I = -\frac{d}{dt}(Q_0 \cos(\omega t)) = -Q_0 (-\omega \sin(\omega t)) = Q_0 \omega \sin(\omega t) \]

### Step 7: Substitute Values
Substituting \( Q_0 = C \cdot V_1 \) and \( \omega = \frac{1}{\sqrt{LC}} \):
\[ I = C \cdot V_1 \cdot \frac{1}{\sqrt{LC}} \sin(\omega t) \]

### Step 8: Express \( \sin(\omega t) \)
Using the identity \( \sin^2(\omega t) + \cos^2(\omega t) = 1 \):
\[ \sin^2(\omega t) = 1 - \cos^2(\omega t) = 1 - \left(\frac{V_2}{V_1}\right)^2 \]
Thus, \( \sin(\omega t) = \sqrt{1 - \left(\frac{V_2}{V_1}\right)^2} \).

### Step 9: Final Expression for Current
Substituting \( \sin(\omega t) \) back into the equation for current:
\[ I = C \cdot V_1 \cdot \frac{1}{\sqrt{LC}} \sqrt{1 - \left(\frac{V_2}{V_1}\right)^2} \]
This simplifies to:
\[ I = \frac{C \cdot V_1}{\sqrt{LC}} \sqrt{1 - \frac{V_2^2}{V_1^2}} \]
Finally, this can be expressed as:
\[ I = \frac{C (V_1^2 - V_2^2)}{L} \]

### Final Answer
The current through the inductor when the potential difference across the capacitor reduces to \( V_2 \) is:
\[ I = \frac{C (V_1^2 - V_2^2)}{L} \]