An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`.
If resistance is removed from the circuit and the value of inductance is doubled, the variation of current with time in the new circuit is

Options

A. `0.52 cos 314 t`
B. `0.52 sin 314 t`
C. `0.52 sin (314 t+ pi//3)`
D. none of these.

Explanation

To solve the problem step by step, we will analyze the given circuit and apply the relevant formulas for an LCR circuit, followed by the modifications made to the circuit.

### Given Data:
- Inductance, \( L = 20 \times 10^{-3} \, \text{H} \)
- Capacitance, \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Resistance, \( R = 50 \, \Omega \)
- Voltage source, \( V(t) = 10 \sin(314t) \)

### Step 1: Determine the Angular Frequency
The angular frequency \( \omega \) can be directly obtained from the voltage source:
\[ \omega = 314 \, \text{rad/s} \]

### Step 2: Calculate the Impedance of the Original Circuit
The impedance \( Z \) of an LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( X_L = \omega L = 314 \times (20 \times 10^{-3}) = 6.28 \, \Omega \)
- \( X_C = \frac{1}{\omega C} = \frac{1}{314 \times (100 \times 10^{-6})} \approx 31.84 \, \Omega \)

Now substituting the values:
\[ Z = \sqrt{50^2 + (6.28 - 31.84)^2} \]
Calculating \( Z \):
\[ Z = \sqrt{2500 + (-25.56)^2} = \sqrt{2500 + 653.6336} \approx \sqrt{3153.6336} \approx 56.14 \, \Omega \]

### Step 3: Calculate the RMS Voltage
The RMS voltage \( V_{rms} \) is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{V} \]

### Step 4: Calculate the RMS Current
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{56.14} \approx 0.126 \, \text{A} \]

### Step 5: Remove Resistance and Double Inductance
When the resistance is removed, \( R = 0 \), and the new inductance \( L' = 2L = 40 \times 10^{-3} \, \text{H} \).

### Step 6: Calculate New Reactance Values
Calculate the new inductive reactance:
\[ X_L' = \omega L' = 314 \times (40 \times 10^{-3}) = 12.56 \, \Omega \]
The capacitive reactance remains the same:
\[ X_C = 31.84 \, \Omega \]

### Step 7: Calculate New Impedance
The new impedance \( Z' \) is:
\[ Z' = \sqrt{0^2 + (X_C - X_L')^2} = |X_C - X_L'| = |31.84 - 12.56| = 19.28 \, \Omega \]

### Step 8: Calculate New Current Amplitude
The new current amplitude \( I_0 \) is given by:
\[ I_0 = \frac{V_0}{Z'} = \frac{10}{19.28} \approx 0.518 \, \text{A} \]

### Step 9: Determine Phase Angle
Since \( X_C > X_L' \), the circuit behaves as capacitive, leading to a phase angle:
\[ \phi = \tan^{-1}\left(\frac{X_C - X_L'}{R}\right) = \tan^{-1}\left(\frac{31.84 - 12.56}{0}\right) = \frac{\pi}{2} \]

### Step 10: Write the Current Equation
The instantaneous current can be expressed as:
\[ I(t) = I_0 \sin(\omega t + \phi) = 0.518 \sin(314t + \frac{\pi}{2}) \]
This can be simplified to:
\[ I(t) = 0.518 \cos(314t) \]

### Final Answer:
The variation of current with time in the new circuit is:
\[ I(t) = 0.518 \cos(314t) \]