An a.c of 1A at a frequency of 800 cycles per second flows through a coil, the inductance of which is 2.5mH…

1. Analyze the Question

We are given:

• Current ($I$) = $1\text{ A}$
• Frequency ($f$) = $800\text{ Hz}$
• Inductance ($L$) = $2.5\text{ mH} = 2.5 \times 10^{-3}\text{ H}$
• Resistance ($R$) = $5\text{ }\Omega$
• We need to find the power absorbed in the coil.

2. Explain the Concept

In an A.C. circuit containing both a resistor and an inductor (an RL circuit), the power is only dissipated by the resistive component. Inductors store energy in a magnetic field and return it to the circuit; they do not consume power.

The formula for power ($P$) in an A.C. circuit is:

Where:

• $I$ is the root-mean-square (r.m.s) current.
• $R$ is the resistance of the coil.

Note: The inductance and frequency are provided to calculate the inductive reactance ($X_L = 2\pi fL$), but since the question asks for the power absorbed by the coil, we only consider the ohmic resistance.

3. Show the Solution

Using the power formula:

Substitute the given values:

The inductive reactance ($X_L = 2\pi fL$) would be $2 \times 3.142 \times 800 \times 0.0025 \approx 12.57\text{ }\Omega$, but this value is irrelevant to the calculation of real power absorbed, as real power is only consumed by the resistance.

4. Identify the Answer

The power absorbed is $5\text{ W}$, which corresponds to option A.

5. Exam Tip

In physics problems, examiners often include extra data (like frequency and inductance) to test if you truly understand which components of a circuit consume power. Always remember: Power is only dissipated by resistors. Inductors and capacitors do not consume real power in an ideal circuit.