JEE Mains Physics · Current Electricity

Electrical Power and Heating

A resistor turns electrical energy into heat at the rate P = VI = I²R = V²/R; pick the form that holds fixed what the circuit holds fixed, current in series and voltage in parallel.

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Why this matters

Thirty PYQs, eighteen of them multiple choice, and three from 2026. Fourteen start from a rating such as 220 V, 100 W: a bulb on a different supply, two bulbs in series, the resistor that lets a lamp run at its rating, a wire cut and reconnected, and the heating times of kettles and coils. Sixteen use H = I²Rt: energy rescaled when the current changes, heat shared among resistors in a circuit, power-line and motor losses, and one that melts ice.

Concept 1 of 2: Power ratings, bulbs and heaters

A rating like 220 V, 100 W tells you one fixed thing: the device's resistance, R = V²/P. Put it in any other circuit and that resistance stays the same while the power changes. In series every device has the same current, so the one with more resistance, the lower-rated bulb, dissipates more and glows brighter. In parallel every device has the same voltage, so the one with less resistance dissipates more.

Definition

From a rating: $R = \dfrac{V_0^{2}}{P_0}$ . On a supply V: $P = P_0\left(\dfrac{V}{V_0}\right)^{2}$ .
In series (same I): $P \propto R$ . In parallel (same V): $P \propto 1/R$ .
To run a device at its rating on a higher supply, add a series resistor that carries the rated current $I_0 = P_0/V_0$ and drops the extra voltage: $R_s = \dfrac{V - V_0}{I_0}$ .
n identical devices in parallel and then in series on the same supply: the powers are in the ratio $n^{2} : 1$ .
Heating a fixed amount of water from a fixed supply: $t \propto R$ . A shorter element has less R and boils sooner.
Two coils that alone take $t_1$ and $t_2$ on the same supply: in series they take $t_1 + t_2$ ; in parallel $\dfrac{t_1t_2}{t_1 + t_2}$ .
Brightness (illumination) is taken to be proportional to the power.

From a rating

R = \frac{V_0^{2}}{P_0}, \qquad P = P_0\left(\frac{V}{V_0}\right)^{2}, \qquad R_s = \frac{V - V_0}{P_0/V_0}

Worked example

A 120 V, 60 W bulb and a 120 V, 40 W bulb are joined in series across 120 V. Find the power in each. Which glows brighter?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q116Moderate

Example 1 · Current Electricity · Electrical Power and Heating

An electrical bulb rated

220\text{ }V,100\text{ }W

, is connected in series with another bulb rated

220\text{ }V

60\text{ }W

. If the voltage across combination is

220\text{ }V

, the power consumed by the

100\text{ }W

bulb will be about

\_\_\_\_

The resistance stays, the power changes

Off its rated supply a device keeps its resistance, not its wattage. Find R from the rating first, then the new power.

In series the lower rating glows more

A 40 W bulb has more resistance than a 60 W bulb of the same voltage, so with the same current it dissipates more. In parallel the order flips.

Power goes as the square

A 10% fall in current or voltage cuts the power by 19%, not 10%. Square the factor.

Drill 13 more on power ratings, bulbs and heaters

Concept 2 of 2: Joule heating, losses and efficiency

Every coulomb that falls through a potential difference V gives up energy V. In a resistor all of it becomes heat, so the heat in a time t is I²Rt. For a resistor with a fixed resistance, doubling the current quadruples the heat. Across a network, the same idea shares heat among resistors: in proportion to R where they share a current, and to 1/R where they share a voltage.

Definition

$H = I^{2}Rt = VIt = \dfrac{V^{2}}{R}t$ . The work done by a source moving charge Q through V is $QV$ .
Same resistor, new current and time: $H \propto I^{2}t$ .
Series resistors: heat $\propto R$ . Parallel resistors: heat $\propto 1/R$ .
"Power in the whole circuit" includes the cell's internal resistance: $P = \varepsilon I$ .
Transmission: the line current is $I = P/V$ , the loss is $I^{2}R_{\text{line}}$ , and the efficiency is the delivered power over the sent power.
A motor: input VI, output = efficiency × input, loss = the rest. $1\ \text{cal} = 4.2$ J.
Heat used to warm and melt: $Q = mc\Delta T + mL$ , then $t = Q/P$ .

Joule's law

H = I^{2}Rt = VIt = \frac{V^{2}t}{R}, \qquad Q = mc\,\Delta T + mL

Worked example

A resistor gives out 400 J in 10 s when it carries 2 A. How much heat does it give out in 4 s when it carries 5 A?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · 2021 Paper 24 · Q7Moderate

Example 2 · Current Electricity · Electrical Power and Heating

Due to cold weather a

1\text{ }m

water pipe of cross-sectional area

1{\text{ }cm}^{2}

is filled with ice at

-10^{\circ}C

. Resistive heating is used to melt the ice. Current of

0.5\text{ }A

is passed through

4k\Omega

resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice

= 3.33 \times10^{5}\text{ }J{\text{ }kg}^{- 1}

, specific heat of ice

= 2 \times10^{3}\text{ }J

kg^{- 1}

and density of ice

=10^{3}\text{ }kg/m^{3}

In parallel the smaller resistor heats more

Resistors in parallel share a voltage, so heat goes as V²/R: the smaller resistor gets the larger share. Using I²R with a single current gets it backwards.

"Whole circuit" includes r

The power of the whole circuit is εI, which counts the heat inside the cell. The power in the external resistors alone is smaller.

Warm before you melt

Ice below 0 °C must first be warmed to 0 °C (mcΔT) and then melted (mL). Leaving out either term gives a time that is too short.

Drill 15 more on joule heating, losses and efficiency

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Power ratings, bulbs and heaters
From a rating
$R = \frac{V_0^{2}}{P_0}, \qquad P = P_0\left(\frac{V}{V_0}\right)^{2}, \qquad R_s = \frac{V - V_0}{P_0/V_0}$
Joule heating, losses and efficiency
Joule's law
$H = I^{2}Rt = VIt = \frac{V^{2}t}{R}, \qquad Q = mc\,\Delta T + mL$

Watch out for (6)

The resistance stays, the power changes
→ Power ratings, bulbs and heaters
In series the lower rating glows more
→ Power ratings, bulbs and heaters
Power goes as the square
→ Power ratings, bulbs and heaters
In parallel the smaller resistor heats more
→ Joule heating, losses and efficiency
"Whole circuit" includes r
→ Joule heating, losses and efficiency
Warm before you melt
→ Joule heating, losses and efficiency

Test yourself on Current Electricity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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