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Current Electricity - Electrical Energy and Power

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electrical Energy is the total work done by a source of EMF in maintaining an electric current in a circuit for a given time tt. It is measured in Joules (JJ).

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Electric Power (PP) is the rate at which electrical energy is dissipated or consumed in an electric circuit. The SI unit is the Watt (WW), where 1W=1J/s1 W = 1 J/s.

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Joule's Law of Heating states that the heat produced in a resistor is directly proportional to the square of the current (I2I^2), the resistance (RR), and the time (tt) for which the current flows.

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For a given voltage VV, power is inversely proportional to resistance (P∝1RP \propto \frac{1}{R}). In parallel circuits, the component with lower resistance consumes more power.

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For a given current II, power is directly proportional to resistance (P∝RP \propto R). In series circuits, the component with higher resistance consumes more power.

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The commercial unit of electrical energy is the kilowatt-hour (kWhkWh), also known as a 'unit'. 1kWh=3.6×106J1 kWh = 3.6 \times 10^6 J.

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Power Transmission: To minimize power loss (Ploss=I2RcP_{loss} = I^2R_c) during transmission over long distances, electricity is transmitted at high voltage and low current, since Ploss=P2RcV2P_{loss} = \frac{P^2 R_c}{V^2}.

📐Formulae

W=Vq=VItW = Vq = VIt

P=Wt=VIP = \frac{W}{t} = VI

P=I2R=V2RP = I^2R = \frac{V^2}{R}

H=I2RtH = I^2Rt

1 kWh=1000 W×3600 s=3.6×106 J1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}

Ptotal (Series)=(1P1+1P2+...)−1P_{total} \text{ (Series)} = \left( \frac{1}{P_1} + \frac{1}{P_2} + ... \right)^{-1}

Ptotal (Parallel)=P1+P2+...P_{total} \text{ (Parallel)} = P_1 + P_2 + ...

💡Examples

Problem 1:

Two bulbs are rated (100W,220V)(100 W, 220 V) and (60W,220V)(60 W, 220 V). Which bulb has greater resistance? If they are connected in series to a 220V220 V source, which one will glow brighter?

Solution:

Resistance is given by R=V2PR = \frac{V^2}{P}. For Bulb 1: R1=2202100=484ΩR_1 = \frac{220^2}{100} = 484 \Omega. For Bulb 2: R2=220260≈806.7ΩR_2 = \frac{220^2}{60} \approx 806.7 \Omega. Thus, the 60W60 W bulb has greater resistance. In series, the current II is the same. Power dissipated is P=I2RP = I^2R. Since R2>R1R_2 > R_1, the 60W60 W bulb dissipates more power and glows brighter.

Explanation:

Brightness depends on the actual power dissipated. In series, power depends directly on resistance (P∝RP \propto R). In parallel, it would be the 100W100 W bulb that glows brighter because P∝1RP \propto \frac{1}{R}.

Problem 2:

An electric motor takes 5A5 A from a 220V220 V line. Determine the power of the motor and the energy consumed in 2h2 h. Also, calculate the remaining energy if 20,000J20,000 J is lost due to friction.

Solution:

Power P=VI=220×5=1100WP = VI = 220 \times 5 = 1100 W. Energy E=P×t=1100W×(2×3600s)=7,920,000JE = P \times t = 1100 W \times (2 \times 3600 s) = 7,920,000 J. To find the useful energy, we subtract the friction loss: 7920000−200007900000\begin{array}{r} 7920000 \\ - 20000 \\ \hline 7900000 \end{array} Useful Energy = 7.9×106J7.9 \times 10^6 J.

Explanation:

Power is the product of voltage and current. Energy is power multiplied by time in seconds. Friction loss is subtracted directly from the total electrical energy consumed to find the net mechanical energy output.