Introduction & AC Voltage Applied to a Resistor

Why the World Chose AC

So far our circuits ran on direct current (dc) — currents that never change direction. But the mains supply in your home varies like a sine function: an alternating voltage, driving an alternating current (ac).

Why does most electrical energy travel as AC?

• AC voltages can be easily and efficiently converted from one value to another by transformers (Section 7).
• Electrical energy can then be transmitted economically over long distances.
• AC circuits have special characteristics exploited in daily devices — tuning a radio uses one of them (resonance, Section 5).

(A pedantic but charming NCERT footnote: 'ac voltage' literally reads 'alternating current voltage' — contradictory, yet universally accepted. We follow the convention.)

The cast of this chapter: a resistor first (this section), then phasors, inductors, capacitors, their grand combination (LCR), resonance, AC power, and finally the transformer.

AC Across a Resistor: In Phase, Ohm Intact

Apply $v = v_m\sin\omega t$ (amplitude $v_m$, angular frequency $\omega$) across a pure resistor R. Kirchhoff's loop rule gives

$$i = \frac{v_m}{R}\sin\omega t = i_m\sin\omega t, \qquad i_m = \frac{v_m}{R}$$

Ohm's law works equally well for AC and DC. Voltage and current reach zero, minima and maxima at the same instants: they are in phase.

Now the subtlety that defines this chapter:

• The current is positive for half the cycle and negative for the other half — its average over a full cycle is zero.
• But Joule heating is NOT zero: $p = i^2R$ depends on $i^2$, which is always positive.

The instantaneous power $p = i_m^2 R\sin^2\omega t$ has average

$$\bar{p} = \frac{1}{2}i_m^2 R$$

using $\langle\sin^2\omega t\rangle = 1/2$ over a cycle.

Key Point: Zero average current but non-zero average power — because power goes as the square. This single line resolves half the conceptual questions ever asked on this section.

RMS Values: Making AC Look Like DC

To write AC power in the DC form $P = I^2R$, define the root mean square (rms) or effective current:

$$I = \sqrt{\langle i^2\rangle} = \frac{i_m}{\sqrt{2}} = 0.707\,i_m$$

and similarly the rms voltage $V = v_m/\sqrt{2}$. Then

$$P = \frac{1}{2}i_m^2R = I^2R, \qquad V = IR, \qquad P = IV = V^2/R$$

— identical in form to the DC equations. That is the whole point of rms values: with them, AC bookkeeping is DC bookkeeping.

The physical meaning: the rms current is the equivalent DC current that would produce the same average power loss in the resistor.

[NEET Important] Household '220 V' is the rms value. The peak is $v_m = \sqrt{2} \times 220 \approx 311$ V. AC meters (ammeters/voltmeters) read rms values by construction.

[JEE Tip] rms is computed by squaring, averaging, then rooting — in that order. For $i = i_m\sin\omega t$ the average of $i$ is 0, of $|i|$ is $2i_m/\pi$, and the rms is $i_m/\sqrt 2$. Three different 'averages' — exams love mixing them up.

Solved Examples

Example 1: The 100 W bulb (NCERT Example 7.1)

A light bulb is rated 100 W for a 220 V supply. Find (a) the bulb's resistance, (b) the peak voltage of the source, (c) the rms current through the bulb.

Solution:

1. (a) $R = \frac{V^2}{P} = \frac{220^2}{100} = 484\ \Omega$.
2. (b) $v_m = \sqrt{2}V = 1.414 \times 220 = 311$ V.
3. (c) From $P = IV$: $I = \frac{100}{220} = 0.454$ A.

Example 2: Reading an AC expression [NEET Numerical]

The mains voltage is $v = 311\sin(100\pi t)$ V. Find the rms voltage and the frequency.

Solution:

1. Peak: $v_m = 311$ V, so $V = v_m/\sqrt 2 = 311/1.414 = 220$ V.
2. Frequency: $\omega = 100\pi$ rad/s, so $\nu = \omega/2\pi = 50$ Hz.
3. The familiar Indian mains: 220 V rms at 50 Hz.

Example 3: Heater by the numbers [JEE Numerical]

A 1000 W heater runs on 220 V rms mains. Find its resistance, the rms current, and the peak current.

Solution:

1. $R = V^2/P = 220^2/1000 = 48.4\ \Omega$.
2. $I = P/V = 1000/220 = 4.55$ A.
3. $i_m = \sqrt 2 I = 1.414 \times 4.55 \approx 6.4$ A — the wiring must tolerate the peak, not just the rms!

Example 4: Three different 'averages' [JEE Numerical]

For the current $i = 2\sin\omega t$ A, find (a) the average over a full cycle, (b) the rms value.

Solution:

1. (a) $\langle\sin\omega t\rangle = 0$ over a cycle, so $\langle i\rangle = 0$.
2. (b) $I = i_m/\sqrt 2 = 2/1.414 = 1.41$ A.
3. Zero average current, yet a perfectly real 1.41 A effective current heating the circuit — squares don't cancel.

Example 5: When does i equal its rms value? [JEE Numerical]

Starting from $i = 0$ at $t = 0$, at what fraction of the period T does $i = i_m\sin\omega t$ first equal its rms value?

Solution:

1. Set $i_m\sin\omega t = i_m/\sqrt 2$: $\sin\omega t = 1/\sqrt 2$.
2. First solution: $\omega t = \pi/4$, i.e. $\frac{2\pi}{T}t = \frac{\pi}{4}$.
3. Answer: $t = T/8$ — one-eighth of a period after the zero crossing.

Example 6: Average vs peak power [NEET Numerical]

An rms current of 5 A flows through a 10 $\Omega$ resistor on AC mains. Find the average power and the peak instantaneous power.

Solution:

1. Average: $P = I^2R = 25 \times 10 = 250$ W.
2. Peak: $p_{max} = i_m^2 R = (\sqrt 2 \times 5)^2 \times 10 = 500$ W.
3. Instantaneous power oscillates between 0 and 500 W; its time average is exactly half the peak — the $\langle\sin^2\rangle = 1/2$ at work.

Example 7: Zero average current, real heating

The average AC current over a cycle is zero. Why does the resistor still get hot?

Solution:

1. Heating goes as $p = i^2R$, and $i^2$ is positive in both half cycles — the positive and negative lobes of $i$ both deposit heat.
2. Averaging i cancels signs; averaging $i^2$ does not: $\langle i^2\rangle = i_m^2/2 \ne 0$.
3. Hence the rms idea: a DC current of $i_m/\sqrt 2$ would heat exactly as much.

Example 8: Why 'effective' is the right word

In what precise sense is the rms current 'equivalent' to a DC current?

Solution:

1. Pass DC of value I through R: power $I^2R$, steady.
2. Pass AC of rms value I through the same R: average power $\langle i^2\rangle R = I^2R$ — the same average power loss.
3. So a 220 V (rms) AC supply lights a bulb exactly as brightly as a 220 V DC supply would. That equivalence is the definition of 'effective'.

Example 9: Why transmit as AC?

Give the main reason electrical energy is generated and distributed as AC rather than DC.

Solution:

1. AC voltages can be stepped up or down easily and efficiently by transformers (which need changing flux — they simply don't work on steady DC).
2. Stepping up the voltage slashes the current for the same power, cutting $I^2R$ transmission losses — economical long-distance transmission.
3. At the consumer's end the voltage is stepped back down to safe values. The full story completes in Section 7.

Example 10: What the meter reads

An AC ammeter in series with a resistor reads 2.0 A. What is the peak current, and what would an ideal DC-averaging meter read?

Solution:

1. AC instruments report rms: $I = 2.0$ A, so $i_m = \sqrt 2 \times 2.0 = 2.83$ A.
2. A meter that truly averaged $i$ over full cycles would read zero — the average AC current vanishes.
3. This is precisely why rms (not the plain average) is the standard for specifying AC quantities.