Electrical Energy and Power — Revision Notes

⚡ 30-Second Revision

Electrical Energy: — $E = VIt = I^2Rt = \frac{V^2}{R}t$. Unit: Joule (J). Practical unit: Kilowatt-hour (kWh).
Electrical Power: — $P = VI = I^2R = \frac{V^2}{R}$. Unit: Watt (W).
Joule's Heating Effect: — Heat produced $H = I^2Rt$.
kWh to Joules: — $1,\text{kWh} = 3.6 \times 10^6,\text{J}$.
Series Circuit Power: — $P propto R$ (for constant $I$). Higher $R$ means more power.
Parallel Circuit Power: — $P propto 1/R$ (for constant $V$). Lower $R$ means more power.
Resistance from Rating: — $R = V_{rated}^2 / P_{rated}$.

2-Minute Revision

Electrical energy is the total work done by current, measured in Joules or kilowatt-hours (kWh). Electrical power is the rate of doing this work, measured in Watts. Key power formulas are $P=VI$, $P=I^2R$, and $P=V^2/R$.

Energy is simply power multiplied by time ($E=Pt$). Remember that $1,\text{kWh} = 3.6 \times 10^6,\text{J}$. Joule's heating effect ($H=I^2Rt$) explains why resistors heat up, forming the basis for heaters and fuses.

For bulbs in series, the one with higher resistance (lower rated power) glows brighter because current is constant ($P propto R$). In parallel, the one with lower resistance (higher rated power) glows brighter because voltage is constant ($P propto 1/R$).

Always convert units to SI (seconds, Watts) for calculations unless kWh is explicitly required. When an appliance operates at a voltage different from its rating, calculate its resistance first, then find the actual power.

5-Minute Revision

Let's consolidate Electrical Energy and Power. Electrical energy ($E$) is the total work done by an electric current, given by $E = VIt$. This can be expanded using Ohm's Law to $E = I^2Rt$ or $E = (V^2/R)t$.

Its SI unit is the Joule (J). For practical household consumption, the kilowatt-hour (kWh) is used, where $1,\text{kWh} = 3.6 \times 10^6,\text{J}$. Electrical power ($P$) is the rate of energy transfer, $P = E/t$.

The fundamental formula is $P = VI$. Using Ohm's Law, we derive $P = I^2R$ and $P = V^2/R$. The SI unit is the Watt (W).

Joule's Heating Effect: When current flows through a resistor, electrical energy is converted into heat. The heat produced is $H = I^2Rt$. This principle is vital for devices like electric heaters, geysers, and fuses. Fuses protect circuits by melting when current exceeds a safe limit.

Power Rating: Appliances have a power rating (e.g., 100W, 220V). This specifies power at the rated voltage. To find the appliance's resistance, use $R = V_{rated}^2 / P_{rated}$. If operated at a different voltage $V_{actual}$, the actual power consumed is $P_{actual} = V_{actual}^2 / R$.

Series vs. Parallel Circuits:

Series: — Current ($I$) is constant. Power $P = I^2R$. So, $P propto R$. The component with higher resistance dissipates more power and glows brighter (e.g., a 60W bulb is brighter than a 100W bulb in series).
Parallel: — Voltage ($V$) is constant. Power $P = V^2/R$. So, $P propto 1/R$. The component with lower resistance dissipates more power and glows brighter (e.g., a 100W bulb is brighter than a 60W bulb in parallel).

Example: A 500W heater runs for 4 hours. Cost of electricity is ₹8/kWh. Calculate the total cost.

1
Convert power to kW: $P = 500,\text{W} = 0.5,\text{kW}$.
2
Calculate energy in kWh: $E = P \times t = 0.5,\text{kW} \times 4,\text{h} = 2,\text{kWh}$.
3
Calculate cost: Cost $= E \times \text{Rate} = 2,\text{kWh} \times ₹8/\text{kWh} = ₹16$.

Prelims Revision Notes

1
Electrical Energy (E): — Work done by electric current. $E = VIt$. Other forms: $E = I^2Rt$ and $E = (V^2/R)t$. SI unit: Joule (J). Commercial unit: Kilowatt-hour (kWh). Conversion: $1,\text{kWh} = 3.6 \times 10^6,\text{J}$. Remember to convert time to seconds for Joules, and power to kilowatts for kWh.
2
Electrical Power (P): — Rate of energy transfer. $P = E/t$. Fundamental formula: $P = VI$. Derived forms using Ohm's Law: $P = I^2R$ and $P = V^2/R$. SI unit: Watt (W). $1,\text{W} = 1,\text{J/s}$.
3
Joule's Heating Effect: — Heat produced in a resistor is $H = I^2Rt$. This is the basis for electric heaters, geysers, toasters, and fuses. Fuse wire has a low melting point and high resistance to melt quickly under excessive current.
4
Power Rating of Appliances: — An appliance rated $P_0$ at $V_0$ has a resistance $R = V_0^2/P_0$. If connected to a different voltage $V_{actual}$, the actual power consumed is $P_{actual} = V_{actual}^2/R$. Note that $P_{actual} propto V_{actual}^2$.
5
Bulb Brightness and Power Dissipation: — Brightness is directly proportional to the actual power dissipated by the bulb.

* In Series Circuits: Current ($I$) is constant. Power $P = I^2R$. So, $P propto R$. The bulb with higher resistance (lower rated power for a given voltage) will glow brighter. * In Parallel Circuits: Voltage ($V$) is constant. Power $P = V^2/R$. So, $P propto 1/R$. The bulb with lower resistance (higher rated power for a given voltage) will glow brighter.

1
Efficiency ($eta$): — $eta = (P_{out} / P_{in}) \times 100$. Useful for problems involving motors or generators.
2
Cost of Electricity: — Calculate total energy in kWh, then multiply by the cost per kWh. Total Cost = Energy (kWh) $imes$ Rate (₹/kWh).

Vyyuha Quick Recall

P-V-I: 'Power is Very Important!' (P=VI) I-Squared-R: 'I squared R is for Heat!' (H=I^2Rt, P=I^2R) V-Squared-R: 'Voltage squared over R, for Parallel Power!' (P=V^2/R)