In a series L-C-R circuit At resonance quality factor is 100. Now value of self inductance doubled and resistance is decreased two fold then find new value of quality factor.

Answer

To solve the problem, we need to find the new quality factor (Q') of a series L-C-R circuit after certain modifications to the inductance (L) and resistance (R).

### Given:
1. Initial quality factor (Q) = 100
2. Inductance (L) is doubled → New inductance (L') = 2L
3. Resistance (R) is decreased to half → New resistance (R') = R/2

### Formula for Quality Factor:
The quality factor (Q) for a series L-C-R circuit at resonance is given by:
\[ Q = \frac{\omega L}{R} \]
Where:
- \( \omega \) is the angular frequency at resonance, given by \( \omega = \frac{1}{\sqrt{LC}} \).

### Step 1: Express Initial Quality Factor
Using the initial quality factor:
\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} = 100 \]

### Step 2: Express New Quality Factor
Now, we need to find the new quality factor (Q'):
\[ Q' = \frac{\omega L'}{R'} \]

### Step 3: Substitute New Values
Substituting \( L' = 2L \) and \( R' = \frac{R}{2} \):
\[ Q' = \frac{\omega (2L)}{\frac{R}{2}} = \frac{2\omega L}{\frac{R}{2}} = \frac{2 \cdot 2\omega L}{R} = \frac{4\omega L}{R} \]

### Step 4: Relate New Quality Factor to Initial Quality Factor
We can relate \( Q' \) to \( Q \):
\[ Q' = 4 \cdot \frac{\omega L}{R} = 4Q \]
Since \( Q = 100 \):
\[ Q' = 4 \cdot 100 = 400 \]

### Conclusion
The new quality factor \( Q' \) after doubling the inductance and halving the resistance is:
\[ \boxed{400} \]