The quality factor of the resonance circuit is given by (symbols have usual meanings)

$1.\frac{{\omega }_{0}L}{R}$
$2. \frac{1}{{\omega }_{0}RC}$
$3. \frac{{\omega }_{0}C }{R}$
$4. Both \left(1\right) \& \left(2\right)$

 

In the given circuit, switch S is closed at t=0. After a long time, the iron rod is withdrawn from the inductor L. Then, the bulb

1. Glow with the same brightness

2. Glows more brightly

3. Gets dimmer

4. Stops glowing

The circuit is in a steady state when the key is at position 1. If the switch is changed from position 1 to position 2, then the steady current in the circuit will be:
  
1. \(E_o \over R\)
2. \(E_o \over 3R\)
3. \(E_o \over 2R\)
4. \(E_o \over 4R\)

LCR series circuit is connected with an alternating emf source. If the peak current and voltage are $l_{0 }and V_0$ respectively then at resonance the value of wattless current is

$1. l_0$
$2. \frac{l_0}{\sqrt{2}}$
$3. \sqrt{2}{l}_0$
$4. zero$

The value of quality factor ( Q-value) of the series resonant circuit with inductance L = 2.0 H, capacitance C= 32 $\mu F$ and resistance R = 10 $\Omega$, is

1. 25

2. 250

3. 0.25

4. 2.5

In the light of the above statements choose the correct answer from the options given below:
 

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R): 
 

In the light of the above statements choose the correct answer from the options given below:
 

Given below are two statements: 

Given below are two statements: