Electrical Energy & Power

From Microscopic Work to Power

In Section 2 we said that the drift velocity of electrons reaches a steady value because the energy gained from the electric field is continuously dissipated to the lattice through collisions. Let us now make that statement quantitative.

Consider a conductor of resistance $R$ carrying a current $I$. Let $dq = I\,dt$ be the charge that flows between two points across which the potential difference is $V$. The work done by the electric field in pushing this charge across is:

$$dW = V\,dq = V\,I\,dt$$

The power delivered to the conductor — the rate at which electrical energy is converted into heat — is therefore:

$$\boxed{P = \dfrac{dW}{dt} = V\,I}$$

Using Ohm's law $V = IR$, we can re-write this in two more forms:

$$\boxed{P = V I = I^{2} R = \dfrac{V^{2}}{R}}$$

This is Joule's law of heating — the heat dissipated in a purely resistive conductor is $P = I^2 R$ watts, and $H = I^2 R\, t$ joules over time $t$.

SI unit of power

The SI unit of power is the watt (W): $1$ W $= 1$ J s$^{-1} = 1$ V $\cdot$ A. Larger units:

• $1$ kilowatt (kW) $= 10^{3}$ W
• $1$ megawatt (MW) $= 10^{6}$ W
• $1$ horsepower (hp) $\approx 746$ W

[Memory hook] When in doubt, pick the formula that uses quantities the question gives you:

• Given $V$ and $I$ → use $P = VI$.
• Given $I$ and $R$ → use $P = I^2R$.
• Given $V$ and $R$ → use $P = V^2/R$.

Energy Balance in a Simple Circuit

Consider a battery of EMF $\varepsilon$ and internal resistance $r$ connected across an external resistor $R$. The steady current is

$$I = \dfrac{\varepsilon}{R + r}.$$

Power delivered by the source (total):

$$P_{\text{source}} = \varepsilon\,I$$

Power dissipated as heat in $R$ (useful):

$$P_R = I^{2} R$$

Power wasted inside the source as heat in $r$:

$$P_r = I^{2} r$$

Energy conservation gives:

$$\varepsilon\, I \;=\; I^{2} R + I^{2} r \;=\; V\,I + I^{2} r,$$

where $V = IR$ is the terminal potential difference across the battery's external terminals.

Terminal voltage vs EMF

From the above:

$$V = \varepsilon - I\,r.$$

So during discharge (current leaving $+$ terminal), $V < \varepsilon$. During charging (current forced into $+$ terminal by an external source), the sign of $I$ reverses and $V = \varepsilon + I\,r > \varepsilon$.

[JEE Tip] In older JEE problems, the terminal-voltage $V$ is often called the "working voltage" of the cell. It is this $V$, not $\varepsilon$, that a voltmeter reads when the cell is supplying current.

Heating Effect of Current — Everyday Applications

The most common everyday manifestation of $P = I^{2}R$ is heat. This simple formula explains a huge range of domestic and industrial appliances:

1. Incandescent lamps

A tungsten filament, heated to about $2500^\circ$C by $I^2R$, glows white-hot and radiates visible light. Only about 5% of the input electrical power emerges as visible light — the rest becomes infrared heat. This is why incandescent bulbs are being replaced by LEDs and CFLs.

2. Electric heaters and iron

Nichrome coils ($\rho \approx 10^{-6}\ \Omega\cdot$m, high melting point, small $\alpha$) are used because:

• High resistivity $\Rightarrow$ reasonable resistance in a small coil.
• High melting point ($\sim 1400^\circ$C) $\Rightarrow$ can glow red without melting.
• Small $\alpha$ $\Rightarrow$ resistance stays nearly constant as it heats up.

3. Fuses

A fuse is a thin wire of an alloy (tin–lead) designed to melt when the current exceeds a safe limit. Since $P = I^2R$, the wire heats up as $I^{2}$: a small over-current produces a disproportionately large heating.

4. Arc welding

An electric arc between two carbon electrodes, carrying very large currents, produces temperatures above $3000^\circ$C — hot enough to melt and fuse metals.

5. Electric kettle

A kettle rated at $2000$ W on a $230$ V mains supply draws $I = P/V = 2000/230 \approx 8.7$ A. The resistance of its heating element is $R = V^2/P = 230^2/2000 \approx 26.5\ \Omega$.

Key point: For a given $V$, a lower-R appliance draws more current and dissipates more power ($P = V^2/R$). This is why domestic wires to high-wattage appliances (geysers, ACs) must be thick — to handle the larger $I$ without overheating.

Why High-Voltage Transmission?

Consider transmitting power $P$ from a generating station over a long cable of resistance $R_\text{line}$ to a consumer at voltage $V$. The current is

$$I = \dfrac{P}{V}$$

and the power lost as heat in the transmission cable is

$$P_{\text{loss}} = I^{2} R_\text{line} = \left(\dfrac{P}{V}\right)^{2} R_\text{line} = \dfrac{P^{2}\, R_\text{line}}{V^{2}}.$$

Two massive observations:

1. $P_{\text{loss}} \propto 1/V^{2}$. Doubling the transmission voltage cuts the loss by a factor of 4!
2. The thicker the cable (smaller $R_\text{line}$), the smaller the loss — but thicker wires are expensive and heavy.

Example of savings

Suppose we need to deliver $P = 100$ kW over a line with $R_\text{line} = 5\ \Omega$.

• At $V = 220$ V: $I = 454$ A, $P_{\text{loss}} = I^2 R = 1.03 \times 10^6$ W $= 1.03$ MW. Loss exceeds delivered power! — the cable would glow red-hot.
• At $V = 22\,000$ V (22 kV, common for city feeders): $I = 4.54$ A, $P_{\text{loss}} = 103$ W. Tiny.
• At $V = 220\,000$ V (220 kV, long-distance transmission): $I = 0.45$ A, $P_{\text{loss}} \approx 1$ W. Negligible!

This is why the grid uses step-up transformers at generating stations and step-down transformers near consumers. Voltage goes up for transmission and comes down for domestic use.

[NEET/JEE Tip] The exam-favourite derivation of this is: "For a given $P$ and line resistance $R_l$, line loss $P_\text{loss} = P^{2} R_l / V^{2}$." Memorise this one-liner — it shows up in many MCQs.

Maximum Power Transfer Theorem

Consider a battery of EMF $\varepsilon$ and internal resistance $r$ connected to a variable load $R$. The power delivered to the load is:

$$P_R = I^{2} R = \left(\dfrac{\varepsilon}{R + r}\right)^{2} R = \dfrac{\varepsilon^{2}\, R}{(R + r)^{2}}.$$

To find when $P_R$ is maximum, set $dP_R/dR = 0$:

$$\dfrac{dP_R}{dR} = \varepsilon^2 \cdot \dfrac{(R+r)^2 - R \cdot 2(R+r)}{(R+r)^4} = \varepsilon^{2} \cdot \dfrac{r - R}{(R+r)^{3}}.$$

This vanishes when $R = r$, and one can check this is a maximum.

$$\boxed{\,P_R^{\max} = \dfrac{\varepsilon^{2}}{4 r}\,}, \quad\text{when } R = r.$$

Interpretation

• When $R \ll r$: current is large but the tiny $R$ absorbs little power (most is wasted in $r$).
• When $R \gg r$: voltage across $R$ is almost $\varepsilon$ but current is tiny.
• The sweet spot is $R = r$: exactly half the source power goes into the load and half into $r$. Efficiency is only 50%.

Application

Audio amplifiers and RF transmitters "impedance-match" their output stage to the load ($R = r$) to get maximum power transfer. Power utilities do NOT try to maximise power — they try to maximise efficiency. They operate with $R \gg r$ so almost all generated power reaches the consumer.

Key point: Maximum power and maximum efficiency are different optimisation problems. For a given source, you cannot have both.

Solved Examples

Example 1: A Light Bulb

A bulb is rated "$60$ W, $230$ V". Find: (a) its resistance, (b) its current at rated conditions, (c) the energy consumed in a 3-hour day.

Solution:

1. Resistance: $$R = \frac{V^2}{P} = \frac{230^2}{60} = \frac{52900}{60} \approx 881.7\,\Omega$$

2. Current: $$I = \frac{P}{V} = \frac{60}{230} \approx 0.261\,\text{A}$$

3. Energy per day: $$E = Pt = 60 \times 3 = 180\,\text{Wh} = 0.18\,\text{kWh}$$

Final Answer: (a) $881.7\,\Omega$ (b) $0.261\,\text{A}$ (c) $0.18\,\text{kWh per day}$

Example 2: Electric Kettle

A $2000$ W, $230$ V kettle heats $1.5$ kg of water from $25^\circ$C to $100^\circ$C. Assuming all heat goes into the water, estimate the time taken. (Specific heat of water $= 4186$ J kg$^{-1}$ K$^{-1}$)

Solution:

1. Heat required: $$Q = mc\Delta T = 1.5 \times 4186 \times 75 = 470925\,\text{J}$$

2. Time: $$t = \frac{Q}{P} = \frac{470925}{2000} \approx 235.5\,\text{s}$$

3. Converting to minutes: $$235.5\,\text{s} \approx 3.9\,\text{min}$$

Final Answer: About $4$ minutes

Example 3: Fuse Wire

A safety fuse is required to protect a $3$ kW heater at $230$ V. What is the minimum suitable rated current of the fuse?

Solution:

1. Operating current: $$I = \frac{P}{V} = \frac{3000}{230} \approx 13.04\,\text{A}$$

2. A fuse must be rated slightly above the normal operating current.

Final Answer: A $15$ A fuse is appropriate

Example 4: Terminal Voltage Drop

A battery of EMF $12$ V and internal resistance $0.5\,\Omega$ is connected to a $5.5\,\Omega$ bulb. Find: (a) the current, (b) the terminal voltage of the battery, (c) the power delivered to the bulb, (d) the power wasted in the battery.

Solution:

1. Current: $$I = \frac{\varepsilon}{R + r} = \frac{12}{5.5 + 0.5} = \frac{12}{6} = 2\,\text{A}$$

2. Terminal voltage: $$V = \varepsilon - Ir = 12 - 2 \times 0.5 = 11\,\text{V}$$

3. Power in bulb: $$P_R = I^2R = 4 \times 5.5 = 22\,\text{W}$$

4. Power in internal resistance: $$P_r = I^2r = 4 \times 0.5 = 2\,\text{W}$$

5. Check: $$\varepsilon I = 12 \times 2 = 24\,\text{W} = 22 + 2$$

Final Answer: (a) $2$ A (b) $11$ V (c) $22$ W (d) $2$ W

Example 5: The kWh Unit

A household uses four $100$ W bulbs for $5$ hours, one $1500$ W geyser for $30$ minutes, and a $200$ W fan for $8$ hours every day. At ₹$8$ per kWh, find the monthly (30-day) electricity bill.

Solution:

1. Daily energy consumption:

• Bulbs: $4 \times 100 \times 5 = 2000$ Wh
• Geyser: $1500 \times 0.5 = 750$ Wh
• Fan: $200 \times 8 = 1600$ Wh
• Total: $4350$ Wh $= 4.35$ kWh

1. Monthly consumption: $$4.35 \times 30 = 130.5\,\text{kWh}$$

2. Bill: $$130.5 \times 8 = 1044$$

Final Answer: Approximately ₹$1044$ per month

Example 6: Maximum Power Transfer

A $6$ V battery with internal resistance $3\,\Omega$ is connected to a variable resistor $R$. Find the value of $R$ at which the power delivered to $R$ is maximum, and compute that maximum power.

Solution:

1. Maximum power transfer occurs when $$R = r = 3\,\Omega$$

2. Maximum power is $$P_{\max} = \frac{\varepsilon^2}{4r} = \frac{36}{12} = 3\,\text{W}$$

3. Check using current: $$I = \frac{\varepsilon}{R+r} = \frac{6}{6} = 1\,\text{A}$$ $$P_R = I^2R = 1^2 \times 3 = 3\,\text{W}$$

Final Answer:

• $R = 3\,\Omega$
• $P_{\max} = 3\,\text{W}$

Example 7: Two Bulbs in Series vs Parallel

Two bulbs rated "$40$ W, $230$ V" and "$100$ W, $230$ V" are connected: (a) in series across $230$ V, (b) in parallel across $230$ V. Which bulb glows brighter in each case?

Solution:

1. Rated resistances: $$R_{40} = \frac{230^2}{40} = 1322.5\,\Omega$$ $$R_{100} = \frac{230^2}{100} = 529\,\Omega$$ So $R_{40} > R_{100}$.

2. Parallel: Each bulb gets full rated voltage, so each dissipates its rated power. Therefore the $100$ W bulb glows brighter.

3. Series: Same current flows through both, and power is $P = I^2R$. The higher-resistance bulb dissipates more power, so the $40$ W bulb glows brighter.

Final Answer:

• In series: the $40$ W bulb glows brighter
• In parallel: the $100$ W bulb glows brighter

Example 8: Why Do We Use High Voltage for Long-Distance Transmission?

Compare the power loss in a transmission line of resistance $4\,\Omega$ delivering $50$ kW at: (a) $400$ V, (b) $40$ kV.

Solution: Using $$P_{\text{loss}} = \frac{P^2R_{\text{line}}}{V^2}$$

1. At $400$ V: $$P_{\text{loss}} = \frac{(50000)^2 \times 4}{400^2} = 62500\,\text{W} = 62.5\,\text{kW}$$

2. At $40$ kV: $$P_{\text{loss}} = \frac{(50000)^2 \times 4}{40000^2} = 6.25\,\text{W}$$

Final Answer:

• At $400$ V: $62.5$ kW loss
• At $40$ kV: $6.25$ W loss

The loss becomes $10{,}000$ times smaller when voltage is increased by a factor of $100$.

Example 9: Change in Power when Resistance Changes

A conductor dissipates $100$ W when connected to a battery. A second conductor of double the resistance is now connected to the same battery instead. What is the new power dissipation?

Solution: For a fixed-voltage source, $$P = \frac{V^2}{R}$$ If the new resistance is $R_2 = 2R_1$, then $$P_2 = \frac{V^2}{2R_1} = \frac{P_1}{2} = 50\,\text{W}$$

Final Answer: $50$ W

Example 10: Heat Generated in a Coil

A $10\,\Omega$ resistor carries a current of $2$ A for $5$ minutes. How much heat is generated?

Solution:

1. Power: $$P = I^2R = 4 \times 10 = 40\,\text{W}$$

2. Time: $$t = 5 \times 60 = 300\,\text{s}$$

3. Heat: $$H = Pt = 40 \times 300 = 12000\,\text{J} = 12\,\text{kJ}$$

Final Answer: $12000$ J

Example 11: Bulb on Half the Rated Voltage

A "$100$ W, $230$ V" bulb is connected to a $115$ V supply. Assuming the resistance remains unchanged, find the power it dissipates.

Solution:

1. Rated resistance: $$R = \frac{230^2}{100} = 529\,\Omega$$

2. New power: $$P = \frac{115^2}{529} = 25\,\text{W}$$

Final Answer: $25$ W

Example 12: Efficiency of Power Delivery

A $6$ V battery with internal resistance $0.5\,\Omega$ supplies a $1.5\,\Omega$ load. Find: (a) the efficiency $\eta = P_R/P_{\text{source}}$, (b) the efficiency when $R$ is doubled to $3\,\Omega$.

Solution:

(a) For $R = 1.5\,\Omega$

$$I = \frac{6}{1.5 + 0.5} = 3\,\text{A}$$ $$P_R = I^2R = 9 \times 1.5 = 13.5\,\text{W}$$ $$P_{\text{source}} = \varepsilon I = 6 \times 3 = 18\,\text{W}$$ $$\eta = \frac{13.5}{18} = 0.75 = 75\%$$

(b) For $R = 3\,\Omega$

$$I = \frac{6}{3 + 0.5} = 1.714\,\text{A}$$ $$P_R = I^2R \approx 1.714^2 \times 3 \approx 8.82\,\text{W}$$ $$P_{\text{source}} = 6 \times 1.714 \approx 10.29\,\text{W}$$ $$\eta = \frac{8.82}{10.29} \approx 85.7\%$$

Also, $$\eta = \frac{R}{R+r}$$ which gives the same result directly.

Final Answer: (a) $75\%$ (b) $85.7\%$