In short

When a current I flows across a potential difference V, the rate at which electrical energy is converted into some other form (heat, light, mechanical motion) is

\boxed{\;P \;=\; V\,I\;}.

For a pure resistor obeying Ohm's law (V = IR), this becomes two equivalent forms:

P \;=\; I^{2}R \;=\; \frac{V^{2}}{R}.

Pick whichever form matches the quantities you know. For an 1800 W, 230 V air-conditioner, the current is I = P/V = 1800/230 \approx 7.83 A; the effective resistance while running is R = V^{2}/P \approx 29.4\ \Omega.

Electrical energy consumed over time t is E = Pt. In SI, P is in watts and t in seconds, so E is in joules. In your BSES or Adani bill, the unit is the kilowatt-hour:

1\ \text{kWh} \;=\; 1000\ \text{W}\times 3600\ \text{s} \;=\; 3.6\times 10^{6}\ \text{J}.

A 100 W ceiling fan running eight hours a day consumes 0.8 kWh/day, about 24 kWh/month, which at the Delhi domestic tariff of roughly Rs 7.75 per unit is Rs 186 a month.

For a cell of EMF \varepsilon and internal resistance r driving current I through an external resistance R:

  • Power delivered by the cell (total, chemistry to electrical): P_\text{total} = \varepsilon I.
  • Power dissipated inside the cell: P_\text{internal} = I^{2}r.
  • Power delivered to the external load: P_\text{ext} = VI = I^{2}R, where V = \varepsilon - Ir is the terminal voltage.

Maximum power theorem: P_\text{ext} is maximised when R = r, and equals \varepsilon^{2}/(4r) at that point. Half the total power is then lost inside the cell; efficiency is exactly 50%.

The BSES Rajdhani electricity bill that arrives at a three-bedroom flat in Dwarka every sixty days is not a mysterious document. In May it reads 1800 units, Rs 13,950 payable. In November the same flat reads 450 units, Rs 3,200 payable. The difference is one number — the daytime temperature in Delhi — and the single piece of equipment that responds to it: a 1.5-ton window AC in the master bedroom, rated 1800 watts. That AC, running six hours a night through the summer, consumes roughly 324 kWh a month all by itself, which is Rs 2,500 of the bill. A ceiling fan at 80 W running twelve hours a day consumes 29 kWh a month — around Rs 225. An LED bulb at 9 W running five hours a day consumes 1.35 kWh a month — Rs 10.50. The bill is not arbitrary. Every watt, every hour, every rupee is accounted for by P = VI and E = Pt.

This chapter derives those two equations from the definition of potential difference, shows you how to translate the SI units (watts, joules, seconds) into the billing units (kilowatts, kilowatt-hours, months), and works out one of the most useful results in all of circuit theory: the maximum-power-transfer theorem, which tells you why a car battery is designed with a small internal resistance and a mobile phone's power amplifier is designed with its output impedance carefully matched to the antenna. By the end, you will be able to pick up your household bill, point at each appliance, and say which fraction of the total it is responsible for, to within a few percent.

Defining electrical power — from the definition of potential difference

Go back to what "potential difference" means. If the potential difference between two points A and B in a circuit is V, then the work done by the electrostatic field on a charge q moving from A to B is

W \;=\; qV. \tag{1}

Why: this is the definition of potential difference. V_A - V_B = W/q means one volt is one joule of work per coulomb of charge. A charge dropping through a potential difference of V volts is handed qV joules by the field — energy that has to go somewhere.

Now consider a steady current I flowing from A to B through some circuit element — a resistor, a motor, a lamp, a battery being charged, whatever. In a time interval \Delta t, the amount of charge that passes through is

\Delta q \;=\; I\,\Delta t. \tag{2}

The energy handed over to (or extracted from) the element in that time is, by (1),

\Delta W \;=\; (\Delta q)\,V \;=\; V I\,\Delta t. \tag{3}

Power is energy per unit time. Divide by \Delta t:

\boxed{\;P \;=\; \frac{\Delta W}{\Delta t} \;=\; V\,I.\;} \tag{4}

Why: this is the master formula for electrical power. It is true for any two-terminal device — resistor, capacitor being charged, battery being drained, LED, motor — whenever a current I flows across a potential difference V. Nothing in the derivation assumed Ohm's law. Equation (4) is the definition of electrical power.

The units: V is in volts (J/C), I is in amperes (C/s). So VI is in joules per second, which is watts. One watt is one joule per second. A 100 W lamp dissipates 100 joules every second.

The three equivalent forms for a resistor

For a resistor obeying Ohm's law, V = IR. Substitute into (4) two different ways:

Form 1 — eliminate V:

P \;=\; V I \;=\; (IR)\,I \;=\; I^{2}R. \tag{5}

Why: if you know the current through a resistor and the resistance, you do not need to measure the voltage separately — P = I^{2}R computes power directly from those two. This form is the one you use when a resistor is in a series chain where the same current I flows through every element.

Form 2 — eliminate I:

P \;=\; V I \;=\; V\cdot\frac{V}{R} \;=\; \frac{V^{2}}{R}. \tag{6}

Why: if you know the voltage across a resistor and the resistance, P = V^{2}/R computes power directly. This form is the one you use when resistors are in parallel across the same voltage source — every branch sees the same V but different V^{2}/R because the resistances differ.

Equations (4), (5), and (6) are three faces of the same physics. Choose by what you know:

If you know... Use
V and I P = VI
I and R P = I^{2}R
V and R P = V^{2}/R

All three give the same number.

A sanity check — the toaster

A pop-up toaster rated 230 V, 1400 W has a nichrome heating element. Three cross-checks using the three forms:

  1. Current from P = VI: I = P/V = 1400/230 \approx 6.09 A.
  2. Resistance from P = V^{2}/R: R = V^{2}/P = 230^{2}/1400 = 52900/1400 \approx 37.8\ \Omega.
  3. Cross-check with P = I^{2}R: P = 6.09^{2}\times 37.8 = 37.1 \times 37.8 \approx 1402 W. ✓

The three forms agree to rounding. You can use whichever is convenient; they are not independent physics, just algebraic rearrangements of the same equation.

Explore the relationship yourself

Before the energy half of the chapter, spend a moment with the three forms. The interactive figure below lets you drag a voltage slider and watch the current and power change for a fixed resistor, or drag a resistance slider at fixed voltage and watch how the heating power of an element scales. The heating element is fixed at 230 V (Indian mains) and you can slide R from 20 Ω (a 2.6 kW geyser rod) up to 200 Ω (a weak 260 W element). The cost-per-hour readout uses the Delhi domestic slab tariff of Rs 7.75 per kWh.

Interactive: power dissipated versus resistance at 230 V A curve P equals V squared over R as a function of R, with R ranging from 20 to 200 ohms at fixed V equals 230 volts. A draggable point on the x-axis selects R and reads out the dissipated power and cost per hour at Rs 7.75 per kWh. resistance R (Ω) power dissipated P (W) 20 60 100 140 180 0 1000 2000 3000 P = V²/R at V = 230 V drag the red point along the R-axis
Drag the red point along the resistance axis. At $R = 20\ \Omega$ (a bath-geyser immersion rod), $P = 2645$ W and the running cost is about Rs 20.50 per hour. At $R = 200\ \Omega$ (a weak heating coil), $P = 265$ W and the cost falls to about Rs 2.05 per hour. Halving the resistance doubles the power — the $1/R$ dependence is sharp.

Electrical energy — from watts to kilowatt-hours

Power is the rate at which energy is converted. Energy is power multiplied by the time it flows:

E \;=\; P\,t. \tag{7}

In SI, P in watts and t in seconds gives E in joules. Fine for physics problems. Useless for your electricity bill.

Why the kilowatt-hour

A joule is tiny. A 100 W bulb, running for one hour, consumes

E \;=\; 100\ \text{W}\times 3600\ \text{s} \;=\; 3.6\times 10^{5}\ \text{J}.

A whole Delhi flat running for a month consumes something like 10^{9} joules — an awkward number to print on a bill. So electricity is billed in a larger unit that matches the scale of actual use: the kilowatt-hour (kWh, also called a unit):

1\ \text{kWh} \;\equiv\; 1\ \text{kilowatt}\times 1\ \text{hour} \;=\; 1000\ \text{W}\times 3600\ \text{s} \;=\; 3.6\times 10^{6}\ \text{J}. \tag{8}

Why: the kWh is the energy a 1 kW appliance consumes in 1 hour. It is the unit the electricity meter in your stairwell clicks through. One kWh is 3.6 megajoules — still a large quantity, but now a Delhi flat uses a reasonable number of them per month (a few hundred) rather than billions of joules.

To go from nameplate watts to monthly kWh, you just multiply watts by hours of use and divide by 1000:

\text{kWh per month} \;=\; \frac{P_\text{W}\;\times\;\text{hours per day}\;\times\;\text{days per month}}{1000}. \tag{9}

A real Delhi flat's monthly consumption

Take a three-bedroom flat in Dwarka in May, Delhi domestic tariff around Rs 7.75/unit (the higher slab that applies above 400 units). Here is the consumption table, built appliance by appliance:

Appliance Power P (W) Hours/day Days/month kWh/month Cost (Rs)
1.5 ton window AC 1800 6 30 324.0 2511
Refrigerator (350 L) 150 (avg) 24 30 108.0 837
Ceiling fans (3 × 75 W) 225 12 30 81.0 628
Geyser (15 L) 2000 0.5 30 30.0 232
LED TV (43 inch) 90 4 30 10.8 84
LED bulbs (8 × 9 W) 72 5 30 10.8 84
Mixer-grinder 500 0.2 30 3.0 23
Washing machine 500 1 10 5.0 39
Microwave oven 1200 0.2 20 4.8 37
Laptop + phone chargers 80 8 30 19.2 149
Total 596.6 4624

That is May. In November, the AC is off, geyser usage drops (warm water is not needed), and the total falls to maybe 200 units — about Rs 1,550 on the lower slab. The summer bill is dominated by one appliance: the air conditioner, contributing 54% of the total. Every other choice — LED bulbs, inverter fridge, efficient fans — matters at the margin; the AC is the main event.

The BEE star rating

The Bureau of Energy Efficiency (BEE) sticker you see on new appliances — the little stars — is a direct reading of E = Pt for a standardised usage pattern. A 5-star 1.5-ton AC consumes about 900 kWh per year. A 3-star 1.5-ton AC consumes about 1200 kWh. The difference, at Rs 7.75/unit, is Rs 2,325 every year — which is why a 5-star AC that costs Rs 8,000 more up-front pays for itself in about 3.5 years.

Power delivered by a cell — chemistry to electricity

A cell is not just a resistor; it is a source. The current it drives comes from chemical energy being converted to electrical energy at a rate set by the EMF and the current.

Recall from the previous chapter: a cell of EMF \varepsilon and internal resistance r driving a current I into an external resistance R obeys

\varepsilon \;=\; I(R + r),\qquad V \;=\; \varepsilon - I r, \tag{10}

where V is the terminal voltage.

Multiply Kirchhoff's loop equation \varepsilon = I(R + r) on both sides by I:

\varepsilon I \;=\; I^{2}R \;+\; I^{2}r. \tag{11}

Equation (11) is an energy budget, per second:

Why: multiplying a voltage by a current gives a power (from equation 4). Equation (10) is a voltage equation — every term has units of volts. Multiplying through by the current I converts each term into a power. The cell sources total power \varepsilon I; the external resistor burns I^{2}R; the internal resistance burns I^{2}r. The chemistry pays for both.

This is why a flashlight's batteries get warm while the bulb is lit: the I^{2}r loss is real and is dissipated as heat inside the cell.

Writing the load power in one variable

Use Ohm's law for the whole loop: I = \varepsilon/(R + r). Substitute into P_\text{ext} = I^{2}R:

P_\text{ext} \;=\; \left(\frac{\varepsilon}{R+r}\right)^{2} R \;=\; \frac{\varepsilon^{2}\,R}{(R+r)^{2}}. \tag{12}

Why: once you fix the cell (\varepsilon and r are given), the only thing you can vary is the external load R. Equation (12) tells you, as a single function of R, how much power actually reaches the load. The numerator grows linearly with R, the denominator grows quadratically; there is a competition, and a maximum hides somewhere in between.

Maximum power transfer — when external R equals internal r

Look at equation (12) at its limits:

Somewhere between these two extremes, P_\text{ext} must be maximum. Find the maximum by setting the derivative with respect to R to zero.

Step 1. Write P_\text{ext}(R) from (12).

P_\text{ext}(R) \;=\; \varepsilon^{2}\,\frac{R}{(R+r)^{2}}.

Step 2. Differentiate using the quotient rule, \frac{d}{dR}\left(\frac{u}{v}\right) = \frac{u'v - u v'}{v^{2}} with u = R, v = (R+r)^{2}.

\frac{dP_\text{ext}}{dR} \;=\; \varepsilon^{2}\cdot\frac{(R+r)^{2} \;-\; R\cdot 2(R+r)}{(R+r)^{4}}.

Why: u' = 1 and v' = 2(R+r) (chain rule on the inside). The quotient rule is the cleanest way to differentiate (12) directly.

Step 3. Factor (R+r) out of the numerator.

\frac{dP_\text{ext}}{dR} \;=\; \varepsilon^{2}\cdot\frac{(R+r)\bigl[(R+r) - 2R\bigr]}{(R+r)^{4}} \;=\; \varepsilon^{2}\cdot\frac{r - R}{(R+r)^{3}}.

Why: (R+r) - 2R = r - R after the obvious simplification. One (R+r) factor cancels between numerator and denominator.

Step 4. Set the derivative to zero.

\frac{r - R}{(R+r)^{3}} \;=\; 0 \quad\Longrightarrow\quad R \;=\; r. \tag{13}

Why: the denominator is always positive (both R and r are positive), so the fraction is zero only when the numerator is zero. That gives R = r. To confirm this is a maximum and not a minimum: to the left of R = r the derivative is positive (the function is rising); to the right, negative (falling). So R = r is the peak.

Step 5. Compute the peak power. Substitute R = r back into (12):

P_\text{ext,max} \;=\; \frac{\varepsilon^{2}\,r}{(r+r)^{2}} \;=\; \frac{\varepsilon^{2}\,r}{4r^{2}} \;=\; \frac{\varepsilon^{2}}{4r}. \tag{14}

Why: with R = r the denominator becomes 4r^{2} and one r cancels with the r in the numerator. Equation (14) is the maximum power a source with EMF \varepsilon and internal resistance r can deliver to any external load.

What is the efficiency at maximum power?

At R = r, the total power generated is \varepsilon I = \varepsilon\cdot \varepsilon/(2r) = \varepsilon^{2}/(2r). Of that, P_\text{ext,max} = \varepsilon^{2}/(4r) reaches the load. The ratio — the efficiency — is

\eta_\text{max-power} \;=\; \frac{P_\text{ext}}{\varepsilon I} \;=\; \frac{\varepsilon^{2}/(4r)}{\varepsilon^{2}/(2r)} \;=\; \tfrac{1}{2} \;=\; 50\%. \tag{15}

Why: at the maximum-power operating point, exactly half the chemical energy ends up in the load and half is dissipated inside the cell. Maximum power is not maximum efficiency — for a battery you want to keep running, you would pick R \gg r and accept lower total power but near-100% efficiency.

This is why your phone's Type-C charger is not operating at maximum power transfer when it tops up a battery. A 20 W charger matched to its own output impedance would waste 20 W inside the charger as heat — the charger would be hotter than the phone. Real chargers use high-efficiency regulation circuits to deliver the power with \eta > 90\%, trading off "maximum power" for "useful battery life per rupee of electricity".

Where maximum-power matching does matter is in radio-frequency engineering — the output stage of a mobile-phone transmitter is deliberately matched (impedance matched, the AC-circuit analog of the theorem above) to the 50 Ω antenna to deliver the most signal out into the air. Here you want maximum power to the antenna, and efficiency is a secondary concern handled by the battery size.

Visualise the peak

The load power P_\text{ext} as a function of R/r is a bell-like curve that peaks at R/r = 1 and falls off on both sides.

Power delivered to external load versus R over rA curve showing P external divided by epsilon squared over r as a function of R over r, with a clear peak at R equals r where the value is one quarter, falling to zero at both zero and infinity.ratio R/rP_ext / (ε²/r)0.51357900.1250.25maximum: P = ε²/(4r) at R = rload = short circuit, P = 0load huge, P → 0
Load power as a function of $R/r$ for a fixed cell. The curve rises from zero at $R = 0$ (short circuit — all the voltage is dropped inside the cell), peaks at $R = r$ with the value $\varepsilon^{2}/(4r)$, and decays to zero as $R\to\infty$ (open circuit — no current to deliver any power).

Worked examples

Example 1: The AC running for one hour in May

A 1.5-ton window AC rated 1800 W, 230 V is switched on for one hour in a Dwarka bedroom in May. Compute (a) the current drawn, (b) the effective resistance while running, (c) the total energy consumed in joules and in kilowatt-hours, and (d) the cost of running it for one hour at the Delhi domestic tariff of Rs 7.75 per kWh.

Circuit diagram for an AC on a 230 V mainsA 230 volt source connected to a resistor labelled 1800 watt AC, with current 7.83 amps flowing through it.230 V1800 W ACR ≈ 29.4 ΩI = 7.83 A →
The AC as a 29.4 Ω load across 230 V mains. The current is 7.83 A and the power dissipated (converted to cooling effect plus heat) is 1800 W.

Step 1. Identify knowns. P = 1800 W, V = 230 V, t = 1 hour = 3600 s.

Why: the nameplate gives power and voltage. The time is what the problem asks about. No other numbers are needed.

Step 2. Current from P = VI.

I \;=\; \frac{P}{V} \;=\; \frac{1800}{230} \;=\; 7.826 \text{ A}.

Why: apply equation (4) rearranged. The AC pulls 7.83 A from the 230 V supply — just under the 10 A rating of a typical wall socket, which is why an AC often gets its own dedicated 16 A socket.

Step 3. Effective resistance from P = V^{2}/R.

R \;=\; \frac{V^{2}}{P} \;=\; \frac{230^{2}}{1800} \;=\; \frac{52900}{1800} \;=\; 29.39\ \Omega.

Why: apply equation (6) rearranged. "Effective resistance" here folds together the compressor's motor windings, the fan motor, and the control electronics — it is the single equivalent R that a DC equivalent of the AC would need to dissipate 1800 W on a 230 V DC supply. The true AC impedance is more complex (there are inductive components), but V^{2}/P captures the active power draw, which is what the meter measures and the bill charges for.

Step 4. Energy in joules.

E_\text{J} \;=\; P\,t \;=\; 1800 \times 3600 \;=\; 6.48\times 10^{6} \text{ J}.

Why: equation (7) with time in seconds. Six and a half million joules — enough energy, in principle, to lift a 6-tonne truck a metre into the air — is what the AC pulls from the grid in just 60 minutes.

Step 5. Energy in kilowatt-hours.

E \;=\; \frac{6.48\times 10^{6}\ \text{J}}{3.6\times 10^{6}\ \text{J/kWh}} \;=\; 1.8\ \text{kWh}.

Or more directly: P in kilowatts times t in hours = 1.8\ \text{kW}\times 1\ \text{h} = 1.8 kWh. Same answer.

Why: the kWh bypasses the awkward joule conversion entirely. Multiply nameplate kW by hours of use — done. This is exactly why the utility picks the kWh.

Step 6. Cost.

\text{Cost} \;=\; 1.8 \text{ kWh}\times 7.75 \text{ Rs/kWh} \;=\; \text{Rs } 13.95.

Result: The AC draws I = 7.83 A, has effective resistance R = 29.4\ \Omega, consumes E = 6.48 MJ = 1.8 kWh in one hour, and costs Rs 13.95 per hour of use at the Delhi tariff.

What this shows: The Rs 13.95 figure is the pocket-money answer to "how much does it cost to keep the AC on for an hour?" Multiply by six hours a night, thirty days, and you get Rs 2,511 — which is exactly the AC's line item on the bill we read at the top of the chapter. The formula P = VI, evaluated for a real appliance, predicts a real rupee figure.

Example 2: Maximum power from a car battery into a starter motor

A Maruti car battery has \varepsilon = 12.6 V and internal resistance r = 0.012\ \Omega (12 mΩ). A starter motor is a low-resistance load. What external resistance R maximises the power delivered to the starter, and what is that maximum power?

Car battery driving a starter motor, equivalent circuitAn EMF of 12.6 volts in series with internal resistance r equals 12 milliohms, connected to external load R representing the starter motor. Current of 525 amps flows around the loop at maximum power when R equals r.ε = 12.6 Vr = 0.012 Ωstarter, RI = ε/(R+r)
The Thevenin-equivalent circuit of the car's cranking system. The battery is $\varepsilon$ in series with $r$; the starter is a load $R$ that the engineer can choose (by designing the motor's winding).

Step 1. Apply the maximum-power condition from (13).

R_\text{opt} \;=\; r \;=\; 0.012\ \Omega.

Why: equation (13) is the answer — the motor's effective resistance while cranking should match the battery's internal resistance. In practice, real starter motors run a little above r to improve efficiency, but the design target is near-match.

Step 2. Compute the current at R = r.

I \;=\; \frac{\varepsilon}{R + r} \;=\; \frac{12.6}{0.012 + 0.012} \;=\; \frac{12.6}{0.024} \;=\; 525 \text{ A}.

Why: when the load is matched to the source, the current is \varepsilon/(2r) — half of the short-circuit current \varepsilon/r = 1050 A. Five hundred amps is why a car battery cable is as thick as your thumb.

Step 3. Apply equation (14) to get the maximum power.

P_\text{max} \;=\; \frac{\varepsilon^{2}}{4 r} \;=\; \frac{12.6^{2}}{4\times 0.012} \;=\; \frac{158.76}{0.048} \;=\; 3307 \text{ W} \approx 3.3 \text{ kW}.

Step 4. Check the efficiency.

Total power: \varepsilon I = 12.6\times 525 = 6615 W. Ratio: 3307/6615 = 0.50 — exactly 50%, as predicted.

Why: equation (15) says maximum-power operation has 50% efficiency. The other 3307 W is burned in the battery's own internal resistance, which is why a car battery gets warm after a long cranking attempt.

Result: The starter's optimal effective resistance is 12 mΩ. At that match, the current is 525 A, the motor receives 3.3 kW of mechanical-converting electrical power, and the battery delivers a total of 6.6 kW with 50% efficiency.

What this shows: A Maruti starter motor is rated around 1 kW mechanical output, which is about 2 kW of electrical input at ~50% motor efficiency. That fits comfortably inside the 3.3 kW peak the battery can deliver — so the matching works, cranking lasts, and the engine fires. If the battery's internal resistance doubles on a cold Shimla morning to 24 mΩ, the peak drops to 12.6^{2}/(4\times 0.024) = 1.65 kW, and the starter can no longer turn the cold engine fast enough to compress cylinder air to ignition. The car refuses to start. This is real physics, measured on a real December morning.

Common confusions

If you came here to read your electricity bill and understand why your AC is expensive, you have everything you need. What follows is for readers who want the DC-to-AC extension (why your mains is an RMS voltage and not a peak voltage), the thermodynamics of the I^{2}r dissipation (why it is always positive, connecting to the second law), and the deeper reading of the maximum-power theorem as an impedance-matching theorem.

RMS voltage — why Indian mains is 230 V, not 325 V

The mains supply at any Indian socket is sinusoidal alternating current:

V(t) \;=\; V_{0}\sin(\omega t),

where the peak voltage V_{0}\approx 325 V and \omega = 2\pi\times 50 rad/s (the Indian grid frequency is 50 Hz). The 230 V on the nameplate is not the peak; it is the root-mean-square (RMS) value:

V_\text{rms} \;=\; \sqrt{\left\langle V(t)^{2}\right\rangle} \;=\; \frac{V_{0}}{\sqrt{2}}.

Why RMS? Because the instantaneous power in a resistor is

p(t) \;=\; \frac{V(t)^{2}}{R} \;=\; \frac{V_{0}^{2}\sin^{2}(\omega t)}{R}.

Averaging over a full cycle:

\langle p\rangle \;=\; \frac{V_{0}^{2}}{R}\,\langle\sin^{2}(\omega t)\rangle \;=\; \frac{V_{0}^{2}}{2R} \;=\; \frac{(V_{0}/\sqrt{2})^{2}}{R} \;=\; \frac{V_\text{rms}^{2}}{R}.

Why: the time-average of \sin^{2} over a full cycle is 1/2. Define V_\text{rms} = V_0/\sqrt{2} and the time-average AC power looks identical to DC power with voltage V_\text{rms}. This is why AC is quoted in RMS — so the same formulas P = V^{2}/R and P = VI work, using RMS values for V and I.

The Indian standard is V_\text{rms} = 230 V, so V_{0} = 230\sqrt{2}\approx 325 V. Your LED bulb's capacitor-input driver sees 325 V peaks sixty times a second, even though the "effective" or RMS voltage is 230 V.

Why I^{2}R is always positive — the link to thermodynamics

The dissipation term I^{2}R is always non-negative because both I^{2} and R are non-negative. Physically, this means a resistor can only absorb electrical energy (converting it to heat) — it cannot generate electrical energy from heat.

This is no coincidence. It is the second law of thermodynamics at work in a circuit. If a resistor could cool down spontaneously and emit electrical energy, you could use that energy to drive a heater and raise another resistor's temperature — net effect, heat flowing from cold to hot without any other change, which the second law forbids. So I^{2}R > 0 in every real resistor, at all times, and the heat always flows outwards.

A Peltier junction (covered in Thermoelectricity) is a different device — it pumps heat with the help of an external power source, but that pump needs outside energy. You cannot build a passive component that converts ambient heat to electrical work.

Impedance matching — the AC generalisation of R = r

The condition R = r for maximum power is, in a DC resistive circuit, all there is. For AC circuits with capacitors and inductors, the generalised version is:

Z_\text{load} \;=\; Z_\text{source}^{*}\quad\text{(complex conjugate)}.

If the source has impedance Z_\text{source} = r + j X (resistance plus reactance), the load for maximum power transfer is Z_\text{load} = r - j X — the reactances are conjugate-matched so they cancel in the series loop, and the resistive parts are equal. This is the central theorem of RF engineering. A 2.4 GHz Wi-Fi antenna driver is designed so that its output impedance matches the conjugate of the 50 Ω antenna impedance; a good impedance match delivers the maximum power into free space, and a bad match (reflection coefficient |\Gamma| > 0) wastes power as standing-wave heat in the feedline.

This is also why Mobile phone antenna design is delicate: a case that covers part of the antenna shifts its impedance, spoiling the match, wasting output power, and reducing signal strength. The R = r rule you derived in Step 4 above is the DC tip of a much larger iceberg.

Why a 5-star appliance matters

A 5-star 1.5-ton AC has an annual consumption of about 900 kWh. A 3-star model of the same nominal cooling capacity consumes about 1200 kWh per year. The P = VI nameplate wattage can be the same — what differs is the duty cycle (how often the compressor runs to maintain the set temperature) and the coefficient of performance (heat removed per watt electrical). Inverter ACs vary the compressor speed continuously; fixed-speed ACs cycle on-off. Over a year, the difference is

\Delta E \;=\; 300\ \text{kWh}\times\text{Rs 7.75/kWh} \;=\; \text{Rs 2325/year}.

At a 10-year appliance life, a 5-star unit saves Rs 23,250 — typically more than the price premium over a 3-star. The BEE star rating is a direct application of E = Pt, averaged over a standardised year of use.

Where this leads next