Q.(a) The peak voltage of an ac supply is $300\ \text{V}$. What is the rms voltage?

Concept: RMS and peak values in AC circuits — for a sinusoidal waveform, the RMS value is $1/\sqrt{2}$ times the peak value.

(a)

Peak voltage $V_0 = 300\ \text{V}$.

RMS voltage is given by $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$.

So $V_{\text{rms}} = \frac{300}{\sqrt{2}} = 150\sqrt{2} \approx 212.1\ \text{V}$.

(b)

RMS current $I_{\text{rms}} = 10\ \text{A}$.

Peak current $I_0 = I_{\text{rms}} \times \sqrt{2}$.

So $I_0 = 10\sqrt{2} \approx 14.14\ \text{A}$.

Why rms and peak are linked by $\sqrt{2}$

When we say "AC voltage" or "AC current" in everyday use, we almost always mean the rms (root-mean-square) value. That’s because rms gives the equivalent DC value that would deliver the same power to a resistor. For a sinusoidal waveform — the standard shape of mains AC — the relationship is fixed:

where $V_0$ and $I_0$ are the peak (maximum instantaneous) values.

The factor $\sqrt{2}$ comes from averaging the square of a sine wave over a cycle. It’s not an approximation — it’s exact for a pure sine wave.

(a) Peak voltage given, find rms voltage

1. The peak voltage is $V_0 = 300\ \text{V}$.

2. The rms voltage is:

3. Rationalise or compute numerically:

So the rms voltage is about $212\ \text{V}$.

(b) rms current given, find peak current

1. The rms current is $I_{\text{rms}} = 10\ \text{A}$.

2. Rearranging the formula:

3. Numerically:

So the peak current is about $14.1\ \text{A}$.

Method: Peak–RMS Conversion for Sinusoidal AC

This method uses the fixed relationship between peak and RMS values for a pure sinusoidal waveform. The key formulas are:

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

Where $V_0$ and $I_0$ are the peak (maximum) values.

(a) Peak voltage → RMS voltage

Step 1: Identify the given peak voltage.

$V_0 = 300\ \text{V}$

Step 2: Apply the conversion formula.

Step 3: Simplify (rationalise if needed).

Step 4: Compute numerical value (exam-ready).

Answer: $\boxed{212.1\ \text{V}}$ (or $150\sqrt{2}\ \text{V}$)

(b) RMS current → Peak current

Step 1: Identify the given RMS current.

$I_{\text{rms}} = 10\ \text{A}$

Step 2: Apply the reverse conversion formula.

Step 3: Compute numerical value.

Answer: $\boxed{14.14\ \text{A}}$ (or $10\sqrt{2}\ \text{A}$)

Why this works (concept check)

• For a sinusoidal AC, the RMS value is the DC equivalent that produces the same average power dissipation in a resistor.
• The factor $\sqrt{2}$ comes from averaging the square of $\sin(\omega t)$ over one cycle.
• Important: These formulas are valid only for pure sine waves — not for square waves, triangular waves, or distorted AC.

Here are the common mistakes students make when solving problems on Power Dissipation in Resistors (specifically for AC RMS and peak values), along with clear ways to avoid each.

Mistake 1: Confusing the formula for RMS voltage and peak voltage

The Mistake

Students often write:

but mix them up — using the wrong one for the given data.

Why it happens

They memorise the formula without understanding the relationship:

• RMS is smaller than peak (since $\frac{1}{\sqrt{2}} \approx 0.707$).
• Peak is larger than RMS (by factor $\sqrt{2} \approx 1.414$).

How to avoid

Always ask: “Is the given value the peak or the RMS?”

• If given peak → divide by $\sqrt{2}$ to get RMS.
• If given RMS → multiply by $\sqrt{2}$ to get peak.

For part (a):

Given $V_0 = 300\ \text{V}$ (peak).

Correct:

Mistake 2: Forgetting to square-root or square incorrectly

The Mistake

Some students write:

Why it happens

They confuse RMS with average power formulas (where $P = \frac{V_{\text{rms}}^2}{R}$).

How to avoid

Remember: RMS is root mean square — the “root” part means you take the square root at the end.

• For a sine wave: $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$ (no extra squaring).
• Only square when calculating power, not when converting peak ↔ RMS.

Mistake 3: Using the same formula for current and voltage incorrectly

The Mistake

Students think:

are different formulas — they are identical in form.

Why it happens

They treat current and voltage as separate “types” of problems.

How to avoid

Understand: For any sinusoidal AC quantity:

It works the same for voltage and current.

For part (b):

Given $I_{\text{rms}} = 10\ \text{A}$.

Correct:

Mistake 4: Forgetting units or writing wrong units

The Mistake

Writing $V_{\text{rms}} = 212.1$ without “V” or writing “A” for voltage.

Why it happens

Rushing through the final answer.

How to avoid

Always include the correct unit:

• Voltage → V (volts)
• Current → A (amperes)
• Power → W (watts)

Mistake 5: Not simplifying $\sqrt{2}$ or leaving answer in improper form

The Mistake

Leaving answer as $\frac{300}{\sqrt{2}}$ without rationalising or approximating.

Why it happens

Some students think it’s acceptable to leave a fraction with a radical in the denominator.

How to avoid

In exams, either:

• Rationalise: $\frac{300}{\sqrt{2}} = 150\sqrt{2} \approx 212.1\ \text{V}$
• Or give decimal to 1–2 decimal places (as per question requirement).

Quick Summary Table

Final correct answers for reference:

1. $V_{\text{rms}} = \frac{300}{\sqrt{2}} \approx 212.1\ \text{V}$
2. $I_0 = 10 \times \sqrt{2} \approx 14.14\ \text{A}$