1MW power is to be delivered from a power station to a town 10km away. One uses a pair of Cu wires of radius 0

NCERT Exemplar · Q22

1\ \text{MW}

(i) power is transmitted at

220\ \text{V}

. Comment on the feasibility of doing this.

(ii) a step-up transformer is used to boost the voltage to

11000\ \text{V}

, power transmitted, then a step-down transformer is used to bring voltage to

220\ \text{V}

. (

\rho_{Cu} = 1.7 \times 10^{-8}

SI unit)

Rajasthan RbseSubjective· 5mImportance★★★★★

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Concept understanding — Power Dissipation in Resistors

Power Dissipation in Resistors

Whenever charge is driven through a resistor, electrical energy is converted into heat. The rate of this conversion is the power dissipated.

Why a Resistor Heats Up

Inside a resistor, drifting electrons repeatedly collide with the vibrating lattice ions. Each collision transfers kinetic energy to the lattice, raising its temperature. The source (battery or AC supply) continually does work to keep the current flowing, and that work reappears as heat. This is Joule heating.

The Power Formulas (DC)

The power delivered to any device carrying current $I$ across a potential difference $V$ is

P = VI

For an ohmic resistor $V = IR$ , so this can be written in three equivalent forms:

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

The SI unit is the watt (W), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .

Which form to use depends on what is fixed:

Series elements share the same current, so $P = I^2 R$ shows the larger resistor dissipates more.
Parallel elements share the same voltage, so $P = V^2/R$ shows the smaller resistor dissipates more.

The total heat produced in time $t$ is $Q = Pt = I^2 R t$ — Joule's law of heating.

Power Dissipation with AC

With alternating current the instantaneous power $p(t) = i^2(t)R$ fluctuates, but a resistor still only dissipates energy (it never returns any). The average power over a cycle is written with root-mean-square values:

P_{\text{avg}} = I_{\text{rms}}^2 R = \frac{V_{\text{rms}}^2}{R} = V_{\text{rms}} I_{\text{rms}}

where for a sinusoid $I_{\text{rms}} = I_m/\sqrt{2}$ and $V_{\text{rms}} = V_m/\sqrt{2}$ . This is precisely why rms values are defined: an AC of rms value $I_{\text{rms}}$ heats a resistor at the same average rate as a steady DC of value $I_{\text{rms}}$ .

Important

A pure resistor has power factor 1 — voltage and current are in phase, so all the power supplied is dissipated. In inductors and capacitors, by contrast, the average dissipated power is zero; energy is only stored and returned.

Worked Example

A $100\ \Omega$ resistor carries a current of $0.5\ \text{A}$ . …

Why this formula?

Power Dissipation in Resistors — Why the Formula Holds

Let's build this from first principles. The goal is to understand why a resistor dissipates power as heat, and how the formula $P = I^2 R$ (and its equivalents) arise naturally.

1. What is "Power" in an Electrical Circuit?

Power is the rate of energy transfer — how much energy is converted from one form to another per unit time.

In a resistor, electrical energy is converted into heat (thermal energy).
The fundamental definition of electrical power is:

P = V \cdot I

where:

$P$ = power (watts, W)
$V$ = voltage across the component (volts, V)
$I$ = current through the component (amperes, A)

Why this definition?

Voltage is energy per unit charge ( $V = \frac{W}{q}$ ), and current is charge per unit time ( $I = \frac{q}{t}$ ). Multiplying them gives energy per unit time — exactly power.

2. How Does a Resistor Behave? — Ohm's Law

A resistor obeys Ohm's Law:

V = I \cdot R

where $R$ is resistance (ohms, $\Omega$ ). This is an empirical law — it describes how real resistors behave: the voltage across them is proportional to the current through them.

3. Deriving the Power Dissipation Formulas

We start with $P = V I$ and substitute Ohm's Law in two ways.

Case A: Express power in terms of $I$ and $R$

Replace $V$ with $I R$ :

P = (I R) \cdot I = I^2 R

Interpretation:

For a fixed resistance, power grows with the square of current.
Doubling current quadruples the heat generated — this is why high currents cause wires to overheat.

Case B: Express power in terms of $V$ and $R$

Replace $I$ with $\frac{V}{R}$ :

P = V \cdot \left( \frac{V}{R} \right) = \frac{V^2}{R}

Interpretation:

For a fixed voltage, power is inversely proportional to resistance.
A low-resistance resistor (like a short circuit) dissipates huge power at a given voltage — that's why short circuits are dangerous.

4. The Physical "Why" — Energy Conversion at the Atomic Level

Why does this energy turn into heat?

Electrons moving through a resistor collide with the atoms of the material.
Each collision transfers kinetic energy from the electron to the atom, making the atom vibrate more — i.e., heating up the resistor.
The rate at which this energy is lost by the electrons (and gained by the lattice) is exactly $P = I^2 R$ .

Key insight:

The $I^2$ term appears because:

More current = more electrons per second. …

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Ohmic loss fraction is $\dfrac{P_{\text{loss}}}{P}=\dfrac{PR}{V^2}$ . At $220\ \text{V}$ it is $\approx 89.5$ (loss far exceeds the power sent — infeasible). At $11000\ \text{V}$ it is $\approx 3.58\times10^{-2}$ (about $3.6\%$ — feasible).

Resistance of the line. The current flows out and back, so the total wire length is

L = 2\times 10\ \text{km}=2\times10^4\ \text{m},\qquad A=\pi r^2=\pi(0.5\times10^{-2})^2=7.85\times10^{-5}\ \text{m}^2.

R=\frac{\rho L}{A}=\frac{(1.7\times10^{-8})(2\times10^{4})}{7.85\times10^{-5}}\approx 4.33\ \Omega.

For a delivered power $P$ at line voltage $V$ , the current is $I=P/V$ and the loss is $I^2R$ , so

\frac{P_{\text{loss}}}{P}=\frac{I^2R}{P}=\frac{PR}{V^2}.

(i) At $V=220\ \text{V}$ :

\frac{P_{\text{loss}}}{P}=\frac{(10^{6})(4.33)}{(220)^2}=\frac{4.33\times10^{6}}{4.84\times10^{4}}\approx 89.5.

The fraction exceeds $1$ : the wires would need to dissipate about $90$ times the power delivered. The required current $I=10^6/220\approx 4.5\times10^{3}\ \text{A}$ is impossibly large for such a wire, and the line drop $IR\approx 1.97\times10^4\ \text{V}$ swamps the supply. Transmission at $220\ \text{V}$ is not feasible.

(ii) At $V=11000\ \text{V}$ : …

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Q. $1\ \text{MW}$ power is to be delivered from a power station to a town $10\ \text{km}$ away. One uses a pair of Cu wires of radius $0.5\ \text{cm}$ for this purpose. Calculate the fraction of ohmic losses to power transmitted if

Concept understanding — Power Dissipation in Resistors

Power Dissipation in Resistors

Why a Resistor Heats Up

The Power Formulas (DC)

Power Dissipation with AC

Worked Example

Why this formula?

Power Dissipation in Resistors — Why the Formula Holds

1. What is "Power" in an Electrical Circuit?

2. How Does a Resistor Behave? — Ohm's Law

3. Deriving the Power Dissipation Formulas

Case A: Express power in terms of $I$ and $R$

Case B: Express power in terms of $V$ and $R$

4. The Physical "Why" — Energy Conversion at the Atomic Level

Method: Comparing Ohmic Transmission Losses at Different Voltages

Steps

Q.1 MW1\ \text{MW}1 MW power is to be delivered from a power station to a town 10 km10\ \text{km}10 km away. One uses a pair of Cu wires of radius 0.5 cm0.5\ \text{cm}0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

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Q. $1\ \text{MW}$ power is to be delivered from a power station to a town $10\ \text{km}$ away. One uses a pair of Cu wires of radius $0.5\ \text{cm}$ for this purpose. Calculate the fraction of ohmic losses to power transmitted if

Unlock everything free for 14 days