Q.A $100\ \Omega$ resistor is connected to a $220\ \text{V}$, $50\ \text{Hz}$ ac supply.

Concept: Power dissipation in a purely resistive AC circuit — the resistor dissipates power exactly as it would under a DC voltage equal to the RMS value.

Step 1 — RMS current

For a resistor, Ohm’s law holds for RMS values:

$I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$.

Step 2 — Substitute values

$I_{\text{rms}} = \frac{220\ \text{V}}{100\ \Omega} = 2.2\ \text{A}$.

Step 3 — Power over a full cycle

In a pure resistor, power is always positive and given by $P = V_{\text{rms}} I_{\text{rms}} = I_{\text{rms}}^2 R$.

$P = (2.2)^2 \times 100 = 4.84 \times 100 = 484\ \text{W}$.

Why this is straightforward

A resistor is the simplest AC load. Unlike an inductor or capacitor, it has no phase difference between voltage and current — the current is exactly in step with the voltage at every instant. That means the instantaneous power $p(t) = v(t)\,i(t)$ is always positive (it never returns energy to the source), and the average power over a cycle is just the same as the DC power you’d get if you used the rms values.

The rms value of an AC quantity is defined precisely so that Ohm’s law and the power formula $P = I^2 R$ work exactly as they do in DC — provided you use rms voltage and rms current. That’s the key insight.

Step-by-step solution

1. Identify the given data

• Resistance: $R = 100\ \Omega$
• Supply voltage (rms): $V_{\text{rms}} = 220\ \text{V}$
• Frequency: $f = 50\ \text{Hz}$ (not needed for a pure resistor — it only matters if there’s reactance)

2. Find the rms current using Ohm’s law

For a resistor, the rms current is simply:

Substitute:

3. Compute the net power consumed over a full cycle

In AC circuits, the average power (or real power) for any element is:

where $\phi$ is the phase angle between voltage and current. For a pure resistor, $\phi = 0^\circ$, so $\cos\phi = 1$.

Thus:

Equivalently, using $P = I_{\text{rms}}^2 R$:

4. Why “over a full cycle” matters

Instantaneous power $p(t) = \frac{V_0^2}{R} \sin^2(\omega t)$ oscillates between $0$ and $2P_{\text{av}}$, but its average over one complete cycle is exactly $P_{\text{av}}$. Since the resistor never stores energy, the net energy dissipated per cycle is $P_{\text{av}} \times T$, where $T = 1/f = 0.02\ \text{s}$.

Method: RMS Power in AC Circuits (Joule Heating Method)

This method uses the RMS (Root Mean Square) approach, which is the standard way to handle power in AC circuits because instantaneous power varies sinusoidally, but average power depends on the RMS values.

Steps

Step 1: Identify given data

• Resistance: $R = 100\ \Omega$
• Supply voltage (RMS): $V_{\text{rms}} = 220\ \text{V}$
• Frequency: $f = 50\ \text{Hz}$ (not needed for this calculation — it cancels out in RMS power)

Step 2: Apply Ohm’s law for RMS values

For a purely resistive AC circuit, the RMS current is:

Substitute:

Step 3: Compute average power over a full cycle

For a resistor, the average power is:

Or equivalently:

Using the simplest form:

Final Answer

• (a) RMS current: $2.2\ \text{A}$
• (b) Net power consumed over a full cycle: $484\ \text{W}$

Why this works: In a pure resistor, voltage and current are in phase, so the instantaneous power $p(t) = v(t) i(t)$ is always positive. The average of $p(t)$ over one cycle equals the product of RMS voltage and RMS current — no need to integrate.

Here are the common mistakes students make with this exact problem, and how to avoid each one.

Mistake 1: Confusing Peak and RMS Values

The Mistake:

Students often take the given $220\ \text{V}$ as the peak voltage ($V_0$) and then calculate current using $I_0 = V_0 / R$.

This leads to an incorrect rms current.

Why it’s wrong:

In standard AC supply notation, $220\ \text{V}$ is the rms voltage ($V_{\text{rms}}$), not the peak. The peak voltage is $V_0 = V_{\text{rms}} \times \sqrt{2} \approx 311\ \text{V}$.

How to Avoid:

Always check the problem statement. If it says “$220\ \text{V}$ AC supply”, treat it as rms unless explicitly stated as “peak” or “maximum”.

Correct approach for part (a):

Mistake 2: Using the Wrong Power Formula

The Mistake:

Students use $P = V_{\text{rms}} \times I_{\text{rms}}$ without considering the power factor, or they use $P = I_0^2 R$ (using peak current).

Why it’s wrong:

For a pure resistor, voltage and current are in phase, so power factor $\cos \phi = 1$.

But if you use peak values, you get peak power, not average power over a cycle.

How to Avoid:

For a resistor in AC, the net power consumed over a full cycle is the same as DC power using rms values:

Correct for part (b):

Mistake 3: Including Frequency in the Calculation

The Mistake:

Students see $50\ \text{Hz}$ and try to use it — for example, by calculating $X_L = 2\pi f L$ (but there’s no inductor) or using time-averaging formulas unnecessarily.

Why it’s wrong:

For a pure resistor, frequency does not affect resistance or power dissipation. The $50\ \text{Hz}$ is a distractor.

How to Avoid:

Recognise that frequency matters only when inductors ($L$) or capacitors ($C$) are present. Here, the circuit is purely resistive — ignore the frequency.

Mistake 4: Forgetting the “Over a Full Cycle” Condition

The Mistake:

Students calculate instantaneous power at a specific time (e.g., at peak voltage) and give that as the answer.

Why it’s wrong:

Instantaneous power in AC varies sinusoidally. The question asks for net power over a full cycle, which is the average power.

How to Avoid:

Remember: For a resistor, average power = $I_{\text{rms}}^2 R$. This already accounts for the full cycle.

Quick Summary Table

Final Answer Check:

1. $I_{\text{rms}} = 2.2\ \text{A}$
2. $P = 484\ \text{W}$