In a sereis L-C-R circuit, `C = 10^(-11)` Farad, `L = 10^(-5)` Henry and R = 100 Ohm, when a constant D.C voltage E is applied to the circuit, the capacitor acquires a charge `10^(-9)C`. The D.C source is replaced by a sinusoidal voltage source in which the peak voltage `E_(0)` is equal to the constant D.C volatge E. At resonance the peak value of the charge acquired by the capacitor will be :

Options

A. `10^(-15)C`
B. `10^(-6)C`
C. `10^(-10)C`
D. `10^(-8)C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given values
We are given:
- Capacitance, \( C = 10^{-11} \) F
- Inductance, \( L = 10^{-5} \) H
- Resistance, \( R = 100 \) Ω
- Charge acquired by the capacitor in DC circuit, \( Q = 10^{-9} \) C

### Step 2: Find the DC voltage \( E \)
In a DC circuit, the voltage across the capacitor can be expressed as:
\[ E = \frac{Q}{C} \]
Substituting the values:
\[ E = \frac{10^{-9}}{10^{-11}} = 10^{2} = 100 \text{ V} \]

### Step 3: Replace DC with AC voltage source
Now, we replace the DC source with a sinusoidal voltage source where the peak voltage \( E_0 \) is equal to the constant DC voltage \( E \). Thus,
\[ E_0 = 100 \text{ V} \]

### Step 4: Calculate the resonant frequency \( \omega_0 \)
At resonance in an L-C-R circuit, the inductive reactance equals the capacitive reactance:
\[ \omega L = \frac{1}{\omega C} \]
This gives us the resonant frequency:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
Substituting the values:
\[ \omega_0 = \frac{1}{\sqrt{10^{-5} \cdot 10^{-11}}} = \frac{1}{\sqrt{10^{-16}}} = 10^{8} \text{ rad/s} \]

### Step 5: Find the maximum current \( I_{max} \)
Using Ohm's law for AC circuits, the maximum current can be calculated as:
\[ I_{max} = \frac{E_0}{R} \]
Substituting the values:
\[ I_{max} = \frac{100}{100} = 1 \text{ A} \]

### Step 6: Calculate the peak charge \( Q_{max} \) at resonance
The peak charge on the capacitor at resonance can be calculated using:
\[ Q_{max} = I_{max} \cdot \frac{1}{\omega_0} \]
Substituting the values:
\[ Q_{max} = 1 \cdot \frac{1}{10^{8}} = 10^{-8} \text{ C} \]

### Final Answer
The peak value of the charge acquired by the capacitor at resonance is:
\[ \boxed{10^{-8} \text{ C}} \]

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