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Electricity - Relation between P, V, I and R

Grade 10CBSE

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

🔑Concepts

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Electric Power (PP) is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit. The S.I. unit of power is the Watt (WW).

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According to Ohm's Law, the potential difference (VV) across the ends of a conductor is directly proportional to the current (II) flowing through it, provided physical conditions remain constant, given by V=IRV = IR.

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The power consumed by a device can be expressed as the product of potential difference and current: P=VIP = VI.

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By substituting Ohm's Law (V=IRV = IR) into the power formula, we derive P=I2RP = I^2R. This relationship shows that power is proportional to the square of the current for a constant resistance, often used to calculate heat loss in series circuits.

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Alternatively, by substituting I=VRI = \frac{V}{R} into the power formula, we derive P=V2RP = \frac{V^2}{R}. This shows that for a constant potential difference (like in household parallel circuits), power is inversely proportional to resistance.

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Electrical Energy (EE) is the product of Power (PP) and Time (tt), expressed as E=P×t=VItE = P \times t = VIt. The commercial unit of energy is kilowatt-hour (kWhkWh).

📐Formulae

V=IRV = IR

P=VIP = VI

P=I2RP = I^2R

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

H=I2RtH = I^2Rt

1 Watt=1 Volt×1 Ampere1\text{ Watt} = 1\text{ Volt} \times 1\text{ Ampere}

💡Examples

Problem 1:

An electric bulb is rated 220 V220\,V and 100 W100\,W. When it is operated on 110 V110\,V, what will be the power consumed?

Solution:

First, calculate the resistance (RR) of the bulb using its rated values: R=V2P=2202100=48400100=484 ΩR = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega Now, calculate the power (P′P') at 110 V110\,V: P′=(V′)2R=1102484=12100484=25 WP' = \frac{(V')^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\,W

Explanation:

The resistance of the filament remains constant regardless of the applied voltage. By finding the resistance from the rated specifications, we can determine the actual power consumed at any other voltage using P=V2RP = \frac{V^2}{R}.

Problem 2:

A current of 0.5 A0.5\,A is drawn by a filament of an electric bulb for 1010 minutes. If the resistance of the filament is 440 Ω440\,\Omega, calculate the power of the bulb and the potential difference.

Solution:

Given: I=0.5 AI = 0.5\,A, R=440 ΩR = 440\,\Omega.

  1. To find Potential Difference (VV): V=IR=0.5×440=220 VV = IR = 0.5 \times 440 = 220\,V
  2. To find Power (PP): P=VI=220×0.5=110 WP = VI = 220 \times 0.5 = 110\,W (Alternatively, P=I2R=0.52×440=0.25×440=110 WP = I^2R = 0.5^2 \times 440 = 0.25 \times 440 = 110\,W)

Explanation:

We use Ohm's Law to find the voltage across the bulb and then apply the power formula P=VIP = VI (or P=I2RP = I^2R) to find the rate of energy consumption.

Relation between P, V, I and R - Revision Notes & Key Formulas | CBSE Class 10 Science