Alternating Current

Inductive reactance: Capacitive reactance: Net reactance: Phase angle:

This is a series LR circuit connected to an AC voltage source. Given values: Inductance H. Resistance . AC voltage (assuming RMS, as is standard for AC voltage values). Frequency . 1. Calculate Inductive Reactance ( ): . . . 2. Calculate Impedance ( ): For a series LR circuit, . . . . (This is a 3-4-5 Pythagorean triple scaled by 5: , , so ). 3. Calculate Current ( ): . . 4. Calculate Phase Difference ( ): The phase difference between voltage and current in a series LR circuit is given by . In an LR circuit, voltage leads current (or current lags voltage). . So, . The current in the circuit is 4 A and the phase difference is . This matches option (c).

Step 1: Calculate the total impedance
The total impedance in a series AC circuit with resistance , capacitive reactance , and inductive reactance is given by: Given:



Substitute the values into the formula:
Step 2: Use Ohm's law to find the current
Now that we know the total impedance , we can use Ohm's law to find the current : Where (the voltage supplied by the source), and (the total impedance). Substitute the values: Therefore, the current through the circuit is .

Given
RMS value
Set to find time when current equals RMS value:


Now, to find time taken to drop from this value to zero, solve for when after this point, which occurs at

If , then s = in angular notation

RMS voltage , impedance

Power =

Step 1: Resonance Condition At resonance, inductive reactance equals capacitive reactance: Step 2: Compute Step 3: Compute Power Factor At resonance, power factor is given by: Thus, the correct answer is and .

To solve for the time taken by the current to change from its peak value to its RMS (root mean square) value in an AC circuit, we can start with the given alternating potential . Here, the peak voltage is 200 V.

The RMS value of an AC voltage is given by:

Substituting the peak voltage, we have:

The current in the circuit at any time is given by Ohm's law:

Where . Thus:

The peak current is 10 A, and the RMS current is:

To find the time when the current changes from to , we equate:

Simplifying gives:

The value corresponds to .

Thus, we have:

Solving for :

Therefore, the correct answer is seconds.