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Question

Which of the following is correct regarding electric power?

The correct answer is

P = VI

Understanding Electric Power Formulas

Electric power is the rate at which electrical energy is transferred by an electric circuit. It represents how much energy is used or produced per unit of time.

The standard unit of electric power in the International System of Units (SI) is the watt (W), which is equal to one joule per second.

Fundamental Relationship of Electric Power

Electric power ($\text{P}$) in a DC circuit can be defined based on the voltage across a component and the current flowing through it. Voltage ($\text{V}$) represents the potential difference (energy per unit charge), and current ($\text{I}$) represents the rate of charge flow (charge per unit time).

The fundamental formula relating electric power, voltage, and current is:

$\text{P} = \text{VI}$

Where:

  • $\text{P}$ is the electric power in watts (W)
  • $\text{V}$ is the voltage across the component in volts (V)
  • $\text{I}$ is the current flowing through the component in amperes (A)

Analyzing the Given Options

Let's examine each of the provided options in the context of the fundamental electric power formula $\text{P} = \text{VI}$.

  1. $\text{P} = \text{V}/\text{I}$: This formula is incorrect for electric power. From Ohm's Law ($\text{V} = \text{IR}$), we know that $\text{R} = \text{V}/\text{I}$, where $\text{R}$ is resistance. So, $\text{V}/\text{I}$ represents resistance, not power.
  2. $\text{P} = \text{I}/\text{V}$: This formula is also incorrect for electric power. $\text{I}/\text{V}$ represents the reciprocal of resistance, which is called conductance ($\text{G} = 1/\text{R}$).
  3. $\text{P} = \text{I}^2\text{V}$: This formula is incorrect for electric power. While $I^2$ is part of another power formula ($P = I^2R$), multiplying it by V instead of R does not yield the correct power value based on fundamental principles or standard derived formulas.
  4. $\text{P} = \text{VI}$: This formula is the correct and fundamental expression for electric power in terms of voltage and current.

Derived Electric Power Formulas using Ohm's Law

Using Ohm's Law ($\text{V} = \text{IR}$), we can derive other useful formulas for electric power:

  • Substitute $\text{V} = \text{IR}$ into $\text{P} = \text{VI}$:

    $\text{P} = (\text{IR})\text{I} = \text{I}^2\text{R}$

    This formula relates power to current and resistance.
  • Substitute $\text{I} = \text{V}/\text{R}$ (derived from Ohm's Law) into $\text{P} = \text{VI}$:

    $\text{P} = \text{V}(\text{V}/\text{R}) = \text{V}^2/\text{R}$

    This formula relates power to voltage and resistance.

Therefore, the common formulas for electric power are $\text{P} = \text{VI}$, $\text{P} = \text{I}^2\text{R}$, and $\text{P} = \text{V}^2/\text{R}$.

Comparing the options with these standard formulas, $\text{P} = \text{VI}$ is clearly a correct formula for electric power.

Common Electric Power Formulas
Formula Variables Description
$\text{P} = \text{VI}$ Power (P), Voltage (V), Current (I) Power in terms of voltage and current.
$\text{P} = \text{I}^2\text{R}$ Power (P), Current (I), Resistance (R) Power dissipated by resistance, useful when current and resistance are known.
$\text{P} = \text{V}^2/\text{R}$ Power (P), Voltage (V), Resistance (R) Power dissipated by resistance, useful when voltage and resistance are known.

Conclusion

Based on the analysis of the fundamental definition and derived formulas for electric power, the formula $\text{P} = \text{VI}$ is correct.

Revision Table: Electric Power Concepts

Key Concepts for Electric Power
Concept Definition/Formula
Electric Power (P) Rate of energy transfer in a circuit. Measured in Watts (W).
Voltage (V) Electric potential difference across components. Measured in Volts (V).
Current (I) Rate of flow of electric charge. Measured in Amperes (A).
Resistance (R) Opposition to electric current flow. Measured in Ohms ($\Omega$).
Ohm's Law $\text{V} = \text{IR}$
Power Formulas $\text{P} = \text{VI}$, $\text{P} = \text{I}^2\text{R}$, $\text{P} = \text{V}^2/\text{R}$

Additional Information: Applications of Electric Power Formulas

Understanding the different electric power formulas is crucial in various electrical engineering and physics applications. For example:

  • Calculating Energy Consumption: Electric bills are based on the total energy consumed, which is power multiplied by time ($\text{Energy} = \text{P} \times \text{time}$). Knowing the power rating ($\text{P}$) of an appliance allows you to calculate its energy use over a period.
  • Designing Circuits: Engineers use power formulas to determine the appropriate voltage and current ratings for components, ensuring they can handle the power without overheating or failing. The $\text{P} = \text{I}^2\text{R}$ formula is particularly useful for calculating power dissipated as heat in resistors.
  • Analyzing Power Dissipation: For components like resistors, power is dissipated as heat. The $\text{P} = \text{I}^2\text{R}$ and $\text{P} = \text{V}^2/\text{R}$ formulas are often used to calculate this heat dissipation, which is important for thermal management.
  • Power Transmission: In power transmission lines, minimizing power loss is critical. Power loss in transmission lines is primarily due to the resistance of the wires and is calculated using $\text{P}_{loss} = \text{I}^2\text{R}_{line}$. This is why power is transmitted at very high voltages (and thus lower currents for a given amount of power, based on $\text{P}=\text{VI}$) to reduce current and minimize $\text{I}^2\text{R}$ losses.

Each formula ($\text{P}=\text{VI}$, $\text{P}=\text{I}^2\text{R}$, $\text{P}=\text{V}^2/\text{R}$) is useful in different situations depending on which electrical quantities (voltage, current, resistance) are known or easiest to measure.

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Important Questions from Power

  1. Which one of the following is the value of 1 KWh of energy converted into joules?

  2. An elevator weighing 400 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used? (g = 9.8 m/s 2and there is no frictional loss)

  3. An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s2 and there is no frictional loss.

  4. Units of power is:

  5. One horse power is equal to

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