Q.In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Exercise 7.3 is a pure inductor and Exercise 7.4 is a pure capacitor — both are purely reactive elements.

For any element the average power over a complete cycle is

In a pure inductor the current lags the voltage by $90^\circ$, and in a pure capacitor it leads by $90^\circ$; in both cases $\phi=90^\circ$, so $\cos\phi=0$. …

The average (net) power delivered to an AC element over a full cycle depends on the phase angle $\phi$ between the voltage and the current:

The factor $\cos\phi$ is the power factor. Only a resistive (in-phase) component absorbs net power; a purely reactive component does not.

1. Exercise 7.3 — a pure inductor

Here the AC source drives a pure inductor (no resistance). The current lags the voltage by exactly $90^\circ$, so $\phi=90^\circ$ and

During one quarter-cycle the current builds up and energy is stored in the inductor's magnetic field; during the next quarter-cycle the current falls and that same energy is handed back to the source. Over a complete cycle the energy borrowed exactly equals the energy returned, so the net power absorbed is zero.

2. Exercise 7.4 — a pure capacitor

Now the source drives a pure capacitor. The current leads the voltage by $90^\circ$, so again $\phi=90^\circ$ and

Energy is stored in the capacitor's electric field as it charges and returned as it discharges, with no net loss over a cycle. …

Method: Instantaneous Power Integration Over a Complete Cycle

This method uses the fundamental definition of average power — the time average of instantaneous power over one complete period.

Steps

1. Write the instantaneous power expression For any circuit element, instantaneous power is:

2. Identify the period $T$ of the AC waveform (usually $T = \frac{2\pi}{\omega}$).

3. Compute the average power over one complete cycle:

4. Substitute the specific voltage and current waveforms for the circuit (e.g., sinusoidal, with phase difference $\phi$).

5. Evaluate the integral — for purely sinusoidal $v$ and $i$:

where $\cos \phi$ is the power factor.

Key Insight for Exercises 7.3 and 7.4

• For a pure resistor ($\phi = 0$): …

Common Mistakes: Net Power Absorption Over a Complete Cycle

Students often stumble on this concept because it blends circuit theory with energy reasoning. Here are the most frequent errors — and how to avoid each.

1. ✗ Mistake: Assuming Power is Always Positive

What students do:

They calculate instantaneous power $p(t) = v(t) i(t)$ and forget that power can be negative (energy returning to source).

Why it’s wrong:

In circuits with inductors or capacitors, energy is stored and released. Over a full cycle, the net power can be zero even if instantaneous power is large.

✓ How to avoid:

Always distinguish between instantaneous power and average power. For a pure L or C, the average power over one cycle is:

Key insight: Energy stored in the first half-cycle is returned in the second half.

2. ✗ Mistake: Forgetting Phase Difference Between Voltage and Current

What students do:

They use $P = V_{\text{rms}} I_{\text{rms}}$ without considering the phase angle $\phi$.

Why it’s wrong:

The correct formula for average power is:

For a pure inductor, $\phi = 90^\circ$ (current lags voltage). For a pure capacitor, $\phi = -90^\circ$ (current leads voltage). In both cases, $\cos \phi = 0$.

✓ How to avoid:

Always check the phase relationship:

• Resistor only: $\phi = 0^\circ$, $\cos \phi = 1$ → power absorbed
• Inductor only: $\phi = 90^\circ$, $\cos \phi = 0$ → zero net power
• Capacitor only: $\phi = -90^\circ$, $\cos \phi = 0$ → zero net power

3. ✗ Mistake: Confusing RMS with Peak Values

What students do:

They plug peak voltage $V_0$ and peak current $I_0$ directly into the power formula.

Why it’s wrong:

Average power uses RMS values:

✓ How to avoid:

Always convert peak to RMS before calculating average power. Remember:

• $V_{\text{rms}} = V_0 / \sqrt{2}$
• $I_{\text{rms}} = I_0 / \sqrt{2}$

4. ✗ Mistake: Thinking "Net Power" Means Instantaneous Power at a Specific Time

What students do:

They pick one instant (e.g., when $v$ and $i$ are both positive) and conclude power is absorbed.

Why it’s wrong:

"Net power over a complete cycle" means average — you must integrate over the full period.

✓ How to avoid:

For a pure L or C, sketch the waveforms. You'll see:

• In one quarter-cycle, energy flows into the element
• In the next quarter-cycle, energy flows back out
• The areas under the power curve cancel → net zero

5. ✗ Mistake: Ignoring the Difference Between Exercises 7.3 and 7.4

What students do: …