Alternating Current

The correct option is (C) : I²R

In A.C. circuit power loss P = V I cos φ P = VI = I²R ( ∵ φ = 0 at resonance)

= tan^-1(3/3)

= tan^-1(1)

= π/4 rad

Hence, option B) π/4 is the correct answer.

Quality Factor and Tuning

For better tuning of an electrical circuit, the quality factor (Q) should be high. The quality factor is a measure of the resonance sharpness of a circuit, and it depends on the resistance , inductance , and capacitance . The formula for the quality factor is given by:

For Maximum Quality Factor:

• For to be maximum, the value of should be minimized.
• A lower resistance leads to a higher quality factor and better tuning quality in the circuit.

Given Values:

The given values for the components are:

Conclusion:

To achieve the best tuning, the resistance should be as low as possible, and with the given values of , , and , the quality factor will be optimized for the best tuning.

Impedance and Power Calculation

The following formulas and calculations are used to determine the impedance and power in an RLC circuit.

1. Reactance of Capacitor :

The reactance of the capacitor is given by:

2. Reactance of Inductor :

The reactance of the inductor is given by:

3. Total Impedance :

The total impedance of the RLC circuit is given by:

Substitute the known values:

4. Power :

To calculate the power, we use the formula:

Substitute the given values:

After simplifying:

Conclusion:

The power dissipated in the circuit is approximately .

To find the angular frequencies at which the power transferred to a series LCR circuit is half of that at the resonant frequency, we can follow these steps:

The resonant angular frequency in a series LCR circuit is given by:

Where is the inductance and is the capacitance.

Given and ,

we calculate as follows:

At resonance, the impedance is purely resistive, so the power transferred at resonance can be expressed as:

Where is the source voltage and is the resistance. Power at the half-power points is:

Impedance at half-power points and the relation with resonance frequency gives the condition:

This leads to a quadratic equation for :

Plugging in the values: and , we solve

or

This simplifies to a quadratic equation leading us to find the roots:

Therefore, the angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are 46 rad/s and 54 rad/s.

The correct answer is 46 rad/s and 54 rad/s.

To solve this problem, we need to apply the concept of a transformer and the relationship between power, voltage, and current. A transformer consists of two coils, known as the primary and secondary coils, and operates on the principle of electromagnetic induction.

Given:

• Primary voltage,
• Secondary voltage,
• Power consumed by the lamp,
• Power losses in the transformer are ignored.

We need to find the current in the primary circuit, .

Solution Steps:

1. First, calculate the current in the secondary circuit, , using the formula for power:

1. Rearranging the formula to solve for gives:

1. Using the formula for an ideal transformer where power in the primary is equal to power in the secondary ( ), we have:

1. Substitute the known values to solve for :

1. Rearrange to find :

Thus, the current in the primary circuit is 0.2 A. Therefore, the correct answer is 0.2 A.

To solve the problem, we start by analyzing the details given for a series LCR circuit. We have:
L (Inductance) = 10 H,
C (Capacitance) = 10 μF = 10 × 10^-6 F,
R (Resistance) = 50 Ω.
The AC source voltage is V = 200 sin(100t) volt, indicating that the angular frequency ω of the source is 100 rad/s.

The resonant frequency of the LCR circuit is given by the formula:

Substituting the values of L and C, we calculate:

Hz.

This means the resonant frequency Hz.

The frequency of the AC source, v, is related to the angular frequency by . We have ω = 100 rad/s, so:

Hz.

Therefore, both the resonant frequency and the frequency of the AC source v are equal,
i.e., Hz.

Hence, the correct answer is:
Hz.

• Inductance
• Capacitance

An ideal transformer conserves power, so the power on the primary side is equal to the power on the secondary side. The power on the secondary side of the transformer can be calculated using the formula for electrical power:

P = VI

where P is power, V is voltage, and I is current. Given values for the lamp are:

• Voltage (V) = 12V
• Power (P) = 60W

From the formula, we can find the current on the secondary side (I_s):

I_s = P/V

Substituting the given values:

I_s = 60W / 12V = 5A

Using the transformer equation and the fact that power is conserved:

• Power on primary side P_p = Power on secondary side P_s
• P_p = V_p × I_p
• P_s = V_s × I_s

Since P_p = P_s, we have:

V_p × I_p = V_s × I_s

Given, primary voltage V_p = 220V. Substitute for known values:

220V × I_p = 12V × 5A

I_p = (12V × 5A) / 220V

I_p = 60 / 220

I_p ≈ 0.27A

Therefore, the current in the primary winding is approximately 0.27 A.