Electrical Energy & Power

From potential energy to heat

Consider a conductor with end points A and B carrying a current $I$ from A to B. Since current flows from A to B, the potential at A exceeds that at B, and the potential difference across AB is $V = V(A) - V(B) > 0$. In a time interval $\Delta t$, an amount of charge $\Delta Q = I\,\Delta t$ travels from A to B. The change in its potential energy is

$$\Delta U_{\text{pot}} = \Delta Q\,[V(B) - V(A)] = -\Delta Q\,V = -I\,V\,\Delta t < 0.$$

If the charges moved without collisions, this lost potential energy would reappear as kinetic energy, $\Delta K = -\Delta U_{\text{pot}} = I\,V\,\Delta t > 0$. In a real conductor, however, the carriers do not accelerate freely; they drift at a steady velocity because of repeated collisions with the ions of the lattice. During those collisions the energy gained from the field is shared with the atoms, which vibrate more vigorously — that is, the conductor heats up. Thus, as NCERT §3.9 states, the energy dissipated as heat in the conductor during the interval $\Delta t$ is

$$\Delta W = I\,V\,\Delta t.$$

The three forms of power

The energy dissipated per unit time is the power dissipated, $P = \Delta W / \Delta t$. From the relation above,

$$P = VI.$$

Using Ohm's law $V = IR$ to substitute, NCERT writes the same power in two further equivalent forms,

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

This is the ohmic loss (power loss) in a conductor of resistance $R$ carrying a current $I$. It is exactly this power that heats the coil of an electric bulb to incandescence, radiating out heat and light. All three expressions are algebraically identical for a pure resistor; they differ only in which two of the three quantities $V$, $I$, $R$ they reference. The art of solving NEET problems quickly is choosing the form that uses the quantities you already know — and, more importantly, the quantity that stays constant across the comparison.

Choosing the right form

The reason a single physical quantity has three formulae is that the proportionality flips depending on what is held fixed. If the current $I$ is the same for every element — the defining property of a series connection — then $P = I^2R$ tells you power is directly proportional to resistance. The largest resistor dissipates the most. If instead the voltage $V$ is common — the defining property of a parallel connection — then $P = V^2/R$ tells you power is inversely proportional to resistance. Now the smallest resistor dissipates the most.

All of this rests on Ohm's law holding for the elements involved; if you are unsure why $V = IR$ lets the three forms collapse into one, revisit the parent relation.

Joule heating: H = I²Rt

Since power is energy per unit time, the heat $H$ produced in a resistor in a time $t$ at constant current is simply the power multiplied by time:

$$H = Pt = I^2 R t.$$

This is Joule's law of heating. The heat produced is proportional to (i) the square of the current $I$, (ii) the resistance $R$ of the conductor, and (iii) the time $t$ for which the current is passed. The squared dependence on current is why a doubling of current quadruples the heating — a key intuition for fuse and heater problems.

When the current is not steady, the heat must be obtained by integrating the instantaneous power over time, $H = \int I^2 R\,dt$. NEET 2016 used exactly this idea: a charge $Q = at - bt^2$ gives a time-varying current $I = dQ/dt = a - 2bt$, and integrating $I^2R$ from $t = 0$ to the instant the current vanishes yields the total heat $a^3R/6b$.

Energy, the joule and the kWh

Power is the rate of energy transfer; the energy itself is $W = Pt$. The SI unit of power is the watt (W), defined as one joule per second, so the SI unit of energy is the joule (J). For everyday consumption the joule is inconveniently small, so electricity is metered in kilowatt-hours (kWh) — the energy used by a 1 kW appliance running for 1 hour. This is the "unit" printed on your electricity bill.

A 60 W lamp on a 220 V supply consumes 60 J of energy each second. Over one hour it consumes 60 Wh, and over 24 hours it consumes $60 \times 24 = 1440$ Wh = 1.44 kWh — that is, roughly 1.4 "units" of energy per day, as the NIOS worked example shows. The conversion $1\ \text{kWh} = (1000\ \text{W})(3600\ \text{s}) = 3.6 \times 10^6\ \text{J}$ is worth committing to memory.

Bulb-rating problems

A bulb (or heater) is stamped with a rated power and a rated voltage, e.g. "100 W, 220 V". These mean the bulb dissipates 100 W only when 220 V is applied across it. Treating the filament as an ohmic resistor of essentially fixed resistance, the rating fixes that resistance through $P = V^2/R$:

$$R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}}.$$

For a 100 W, 220 V bulb, $R = 220^2 / 100 = 484\ \Omega$; for a 60 W, 220 V bulb, $R = 220^2 / 60 \approx 807\ \Omega$. The crucial consequence: at the same rated voltage, a higher-wattage bulb has a lower resistance. Once the resistance is known, the bulb behaves like any resistor, and its actual power in a circuit depends on how it is wired.

Power delivered & transmission

The power $P = I^2R = V^2/R$ delivered to a resistor depends on the circuit driving it. A real source has its own internal resistance, and the power delivered to an external load is maximised — the maximum-power-transfer condition — when the load resistance equals the source's internal resistance. This is useful background, though NEET more often tests the dissipation forms directly.

Equation $P = V^2/R$ also explains high-voltage transmission. To deliver a power $P$ down cables of resistance $R_c$, the wasted power is $P_c = I^2 R_c = (P/V)^2 R_c$, so the loss in the connecting wires is inversely proportional to $V^2$. Because transmission cables run for hundreds of miles and have considerable $R_c$, electricity is carried at enormous voltages to keep $P_c$ small — the reason for the high-voltage danger signs on transmission lines. A transformer then steps the voltage back down to a safe value for use.