Alternating Current

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :

1. Statement I is true but Statement II is false
2. Statement I is false but Statement II is true
3. Both Statement I and Statement II are true
4. Both Statement I and Statement II are false

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 .

1. 10 A
2. 10 mA
3. 68 A
4. 63 A

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 :

1. A, B and C only
2. A and B only
3. A, B and D only
4. B, C and D only

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:

1. A-IV, B-I, C-III, D-II
2. A-I, B-IV, C-II, D-III
3. A-IV, B-I, C-II, D-III
4. A-I, B-IV, C-III, D-II

In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be:

1. increased by 2 L
2. reduced by 14 L
3. reduced by 34 L
4. increased to 4 L

A transformer has an efficiency of 80\% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is :

1. 1.33 A
2. 13.33 A
3. 1.59 A
4. 15.1 A