A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
(4) double
At resonance XL = XC ⇒ Z = R
I0 = V0R
R R2 (resonance unchanged, 0 = 1 LC) ⇒ I0' = V0R/2 = 2I0
Answer: (4)
A coil of negligible resistance is connected in series with 90 resistor across 120 V, 60 Hz supply. A voltmeter reads 36 V across resistance. Inductance of the coil is :
An alternating emf E = 110 2 100t volt is applied to a capacitor of 2 , the rms value of current in the circuit is mA.
Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :
When a dc voltage of 100 V is applied to an inductor, a dc current of 5 A flows through it. When an ac voltage of 200 V peak value is connected to inductor, its inductive reactance is found to be 20 3 . The power dissipated in the circuit is W.
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 and 100 mH inductor is being used in the given circuit. The frequency of ac source is 103 Hz (given 2 = 10)
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20 is V. [figure]
A series LCR circuit is subjected to an ac signal of 200 V, 50 Hz. If the voltage across the inductor (L = 10 mH) is 31.4 V, then the current in this circuit is .
In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to : Choose the correct answer from the options given below :
An alternating current at any instant is given by i= [6+ 56 (100 t+ /3) ] A. The rms value of the current is A.
Match List I with List II Choose the correct answer from the options given below: