Energy, Power And Heating Effect

PhysicsCurrent ElectricityFor NEET aspirants

When current flows through a resistor, electrical energy is dissipated as heat — a phenomenon called Joule's heating effect. The rate of energy dissipation (power) is given by $P = V I = I^{2} R = V^{2} / R$ , and the total heat produced in time $t$ is $H = I^{2} R t$ joules. The maximum-power transfer theorem states that maximum power is delivered to a load when its resistance equals the source's internal resistance ( $R = r$ ). These formulas underpin electrical bill calculation, fuse ratings, and appliance design, and are core to the JEE Physics - Current Electricity and NEET Physics - Current Electricity syllabi.

Key Formulas — Quick Reference

Electrical energy: $W = Q V = V I t$
Electrical power: $P = V I = I^{2} R = \frac{V ^{2}}{R}$
Joule's heating: $H = I^{2} R t$ joule $= \frac{I ^{2} R t}{4.18}$ calorie
Cell energy balance: $ε I = I^{2} R + I^{2} r$
Maximum power theorem: $P_{m a x}$ at $R = r$ , and $P_{m a x} = \frac{ε ^{2}}{4 r}$
Efficiency of a cell: $η = \frac{R}{R + r} \times 100%$
Power rating of appliance: $P_{rated} = \frac{V _{rated}^{2}}{R}$
Bulbs in series: brighter bulb has higher $R$ ; bulbs in parallel: brighter bulb has lower $R$
Energy consumed (commercial): 1 kWh = $3.6 \times 1 0^{6}$ J = 1 'unit' on electricity bill

1. Electrical Energy and Power

When a current $I$ flows for time $t$ through a device with potential difference $V$ across it, the charge that flows is $Q = I t$ and the energy delivered by the source is:

W = Q V = V I t

The rate of energy transfer (power) is:

P = \frac{W}{t} = V I

For an ohmic resistor with $V = I R$ :

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

SI units: energy in joule (J), power in watt (W) = J/s. Commercial unit of energy is kilowatt-hour (kWh):

1 kWh = 1 0^{3} W \times 3600 s = 3.6 \times 1 0^{6} J

2. Joule's Heating Effect

When current passes through a conductor, the drifting electrons collide with lattice ions and transfer kinetic energy to them. This raises the temperature of the conductor — the phenomenon is called Joule heating (or ohmic heating).

The heat produced in time $t$ is:

H = I^{2} R t joule

Or equivalently: $H = V I t = \frac{V ^{2} t}{R}$ . In calories: $H = \frac{I ^{2} R t}{4.18}$ .

Laws of Joule Heating

$H \propto I^{2}$ (at fixed $R$ and $t$ ).
$H \propto R$ (at fixed $I$ and $t$ ) — only when current is fixed.
$H \propto t$ (at fixed $I$ and $R$ ).

Important caveat: If voltage is fixed instead of current, then $H = V^{2} t / R$ , so $H \propto 1/ R$ . In series (same $I$ ), a higher $R$ dissipates more heat. In parallel (same $V$ ), a lower $R$ dissipates more heat.

Applications of Joule Heating

Electric heater, iron, geyser: nichrome coils with high resistivity dissipate large heat.
Incandescent bulb: tungsten filament glows white-hot at ~2500 K.
Fuse wire: tin-lead alloy with low melting point; melts and breaks the circuit if current exceeds rated value.
Electric arc welding: intense heating at points of contact.

3. Energy Balance in a Circuit

Consider a cell of EMF $ε$ and internal resistance $r$ connected to an external resistance $R$ . The current is $I = ε / (R + r)$ . Multiplying by $I$ :

ε I = I^{2} R + I^{2} r

Interpretation:

$ε I$ = rate at which chemical energy is converted to electrical energy inside the cell.
$I^{2} R$ = power delivered to (dissipated in) the external load.
$I^{2} r$ = power dissipated inside the cell (heats the cell).

Efficiency of the cell (fraction of energy delivered to load):

η = \frac{power to load}{total power} = \frac{I ^{2} R}{ε I} = \frac{I R}{ε} = \frac{R}{R + r}

4. Maximum Power Transfer Theorem

For a given cell (fixed $ε$ , $r$ ), what value of external resistance $R$ absorbs the maximum power?

Power delivered to $R$ :

P = I^{2} R = \frac{ε ^{2} R}{( R + r ) ^{2}}

To maximise, set $d P / d R = 0$ :

\frac{d P}{d R} = ε^{2} \cdot \frac{( R + r ) ^{2} - 2 R ( R + r )}{( R + r ) ^{4}} = 0

\Rightarrow (R + r) - 2 R = 0 \Rightarrow R = r

The power output is maximum when the external resistance equals the internal resistance.

Substituting back:

P_{m a x} = \frac{ε ^{2} r}{( 2 r ) ^{2}} = \frac{ε ^{2}}{4 r}

Efficiency at maximum power transfer:

η_{m a x P} = \frac{R}{R + r} = \frac{r}{2 r} = 50%

Figure 1: Power delivered to external load

R

peaks at

R = r

with maximum value

ε^{2} /4 r

. Efficiency at this point is only 50%; as

R

grows, efficiency climbs to 100% but the absolute power delivered falls back to zero.

Practical Trade-off

At $R = r$ , power delivered to the load is maximum but only half the total power reaches the load — the other half is wasted heating the cell. For efficient power transmission (e.g. grid), we want $R ≫ r$ so efficiency approaches 100%, even though absolute delivered power is smaller. Maximum power theorem is useful for matching audio impedance and low-power sensors, not for power generation.

5. Bulbs in Series vs Parallel

Consider two bulbs with power ratings $P_{1}$ and $P_{2}$ at rated voltage $V_{0}$ . Their resistances are:

R_{1} = \frac{V _{0}^{2}}{P _{1}}, R_{2} = \frac{V _{0}^{2}}{P _{2}}

Lower power rating means higher resistance (for the same voltage rating).

Series Connection (same current)

With both bulbs in series across supply $V$ , current $I$ is common. Actual power dissipated in each: $P_{i} = I^{2} R_{i}$ . Since $R$ is inversely proportional to rated power:

\frac{P _{1}^{actual}}{P _{2}^{actual}} = \frac{R _{1}}{R _{2}} = \frac{P _{2}^{rated}}{P _{1}^{rated}}

In series, the bulb with the lower power rating glows brighter (because it has higher resistance).

Parallel Connection (same voltage)

With both bulbs in parallel across supply $V = V_{0}$ : $P_{i} = V_{0}^{2} / R_{i}$ . Directly:

\frac{P _{1}^{actual}}{P _{2}^{actual}} = \frac{R _{2}}{R _{1}} = \frac{P _{1}^{rated}}{P _{2}^{rated}}

In parallel, the bulb with the higher power rating glows brighter (household wiring uses parallel so each appliance runs at rated brightness).

Solved Example 1

Q: A copper wire has cross-section $0.5$ mm $^{2}$ and length $0.1$ m, initially at $2 5^{\circ}$ C, thermally insulated. A current of $10$ A is set up. (a) Find the time in which the wire starts melting. (b) What if the length is doubled?

Given: density of Cu $= 9 \times 1 0^{3}$ kg/m $^{3}$ , specific heat $= 9 \times 1 0^{- 2}$ cal/(kg $\cdot^{\circ}$ C) $\times 4.2$ J/cal, melting point $= 107 5^{\circ}$ C, resistivity $= 1.6 \times 1 0^{- 8} Ω \cdot$ m. Neglect temperature variation of $R$ .

Solution

(a) Mass of wire:

m = V \times ρ_{Cu} = (0.5 \times 1 0^{- 6}) (0.1) (9 \times 1 0^{3}) = 4.5 \times 1 0^{- 4} kg

Temperature rise: $Δ T = 1075 - 25 = 105 0^{\circ}$ C.

Specific heat: $s = 9 \times 1 0^{- 2} \times 4.2 = 0.378$ J/(kg $\cdot^{\circ}$ C).

Resistance:

R = \frac{ρ L}{A} = \frac{1.6 \times 1 0 ^{- 8} \times 0.1}{0.5 \times 1 0 ^{- 6}} = 3.2 \times 1 0^{- 3} Ω

Setting heat produced equal to heat required to melt:

I^{2} R t = m s Δ T

t = \frac{m s Δ T}{I ^{2} R} = \frac{( 4.5 \times 1 0 ^{- 4} ) ( 0.378 ) ( 1050 )}{( 100 ) ( 3.2 \times 1 0 ^{- 3} )} \approx 558 s

(b) If length is doubled: $R$ doubles, but so does $m$ . The ratio $m R$ (which appears when substituting $R = ρ L / A$ into the formula) leaves $t$ unchanged. Wire melts in the same time.

Solved Example 2

Q: A 100 W, 220 V bulb and a 60 W, 220 V bulb are connected (a) in series across 220 V, (b) in parallel across 220 V. In each case, which bulb glows brighter?

Solution

Resistances: $R_{100} = 22 0^{2} /100 = 484 Ω$ , $R_{60} = 22 0^{2} /60 = 806.7 Ω$ .

(a) Series: total $R = 1290.7 Ω$ , current $I = 220/1290.7 = 0.170$ A.

P_{100}^{actual} = I^{2} R_{100} = (0.170)^{2} (484) = 14.0 W

P_{60}^{actual} = I^{2} R_{60} = (0.170)^{2} (806.7) = 23.3 W

The 60 W bulb glows brighter in series (higher $R$ ).

(b) Parallel: each bulb sees full 220 V, so each dissipates its rated power. The 100 W bulb glows brighter (higher rated power).

Solved Example 3

Q: An electric geyser rated 2000 W, 220 V is used for 30 minutes daily. If electricity costs ₹8 per unit (kWh), find the monthly (30-day) cost of running the geyser.

Solution

Daily energy: $E = P \times t = 2000 W \times 0.5 h = 1000 Wh = 1 kWh$ .

Monthly energy: $30 \times 1 = 30$ kWh = 30 units.

Cost: 30 \times 8 = \rupee\,240 per month.

Fuse Wire Design

A fuse is a thin wire with low melting point (tin-lead alloy) placed in series with the load. Its cross-section is chosen so that Joule heating melts it once current exceeds the safe rated value. Since heat produced per unit length $\propto I^{2}$ , and heat loss depends on surface area, the fuse rating scales as $I \propto d^{3/2}$ where $d$ is diameter.

Frequently Asked Questions

Why does maximum power transfer happen at

R = r

? (JEE Main / NEET)

Power delivered to load is

P = ε^{2} R / (R + r)^{2}

. If

R

is very small, the current is high but very little voltage drops across

R

, so

P

is small. If

R

is very large, voltage across

R

approaches

ε

but current is very small, so

P

is again small. The peak occurs at the balance point

R = r

, giving

P_{m a x} = ε^{2} /4 r

. Efficiency there is exactly 50%.

Why does a bulb often blow when it is switched on rather than in the middle of use? (JEE / NEET)

When cold, the tungsten filament has a much lower resistance than at operating temperature (metals have positive

α

). At the moment of switch-on, the instantaneous current

I = V / R_{cold}

is very large, producing a sudden thermal shock. This surge current, combined with mechanical stresses on the aged filament, is most likely to cause failure at that instant.

What is the difference between kilowatt and kilowatt-hour? (JEE Main / NEET)

Kilowatt (kW) is a unit of power — the rate at which energy is used or produced (1 kW = 1000 J/s). Kilowatt-hour (kWh) is a unit of energy — the amount of energy consumed in 1 hour at a power of 1 kW. On electricity bills, 1 unit = 1 kWh =

3.6 \times 1 0^{6}

J. So a 100 W bulb burning for 10 hours uses 1 kWh = 1 unit.

If two bulbs of the same wattage are connected one in series and one in parallel to the same source, will they glow equally bright? (JEE Main / NEET)

No. In parallel, each bulb gets the full rated voltage and glows at its rated power. In series, the bulbs share the voltage, so each gets half the rated voltage and dissipates only

P /4

(since

P \propto V^{2}

). Two identical bulbs in series glow equally to each other, but each is much dimmer than the same bulb in parallel.

Previous year questions on Energy, Power And Heating Effect

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

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