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Physics

Measurement of electrical power

Chapter: Electricity and Magnetism

Question 1 of 5 NDA MCQ

Which one of the following correctly representsthe SI unit of resistivity?

Detailed Explanation

Correct Answer

✓ Option (d) is correct. The SI unit of resistivity is ohm-meter (Ω m).

📖 Governing Law / Core Concept

The resistivity (\( \rho \)) of a material is defined by the formula:

\[ \rho = R \cdot \frac{A}{L} \]

where:

  • R: Resistance in ohms (Ω)
  • A: Cross-sectional area in square meters (m²)
  • L: Length in meters (m)

🧪 Step-by-Step Breakdown

From the formula:

\[ \text{Resistivity} (\rho) = R \cdot \frac{A}{L} \]

The unit of resistance (R) is ohms (Ω), area (A) is in square meters (m²), and length (L) is in meters (m). Thus:

\[ \rho \quad \text{(in SI units)} = \Omega \cdot \frac{m^2}{m} = \Omega \cdot m \]

Therefore, the SI unit of resistivity is ohm-meter (Ω m).

🔍 Option Analysis

  • Option A (Ω): Represents resistance, not resistivity.
  • Option B (Ω/m): Incorrect as it implies resistance per unit length.
  • Option C (Ω cm): Represents resistivity in a specific non-SI unit.
  • Option D (Ω m): Correct as it represents the SI unit of resistivity.

⚡ Mnemonic / Speed-Run

Remember: "Resistance per Length gives Resistivity" (Ω m).

6. Visual Suggestion

6. Visual Suggestion

Visualize a simple circuit diagram with a resistor labeled with its value in ohms, along with a formula sheet displaying the relationship \( \rho = R \cdot \frac{A}{L} \) to illustrate how resistivity is derived from resistance.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 3 (Current Electricity).

Question 2 of 5 NDA MCQ

The commercial unit of electrical energy is kilowatt–hour (kWh), which is equal to

Detailed Explanation

Correct Answer

✓ Option (a) is correct. 1 kWh is equal to 3.6 × 106 J.

📖 Governing Law / Core Concept

The commercial unit of electrical energy is defined by the relationship:

\[ 1 \, \text{kWh} = 1000 \, \text{W} \times 3600 \, \text{s} \]

1 kWh represents the energy consumed by a device operating at 1 kW for 1 hour.

🧪 Step-by-Step Breakdown

\[ \begin{aligned} 1 \, \text{kWh} &= 1000 \, \text{W} \times 3600 \, \text{s} \\ &= 1000 \times 3600 \, \text{J} \\ &= 3.6 \times 10^6 \, \text{J} \end{aligned} \]

Where:

  • Power (P) = 1000 W
  • Time (t) = 3600 seconds (1 hour)
  • Energy (E) = Joules (J)

🔍 Option Analysis

Other options are incorrect based on the energy conversion:

  • B: 3.6 × 103 J - Significantly lower than kWh.
  • C: 103 J - Not equivalent to kWh.
  • D: 1 J - Same as the previous calculations; negligible for a kWh conversion.

⚡ Mnemonic / Speed-Run

  • Remember, 1 kWh = 1000 W × 3600 s. Use "KiWi" for Kilo Watt i.e., 1 Q for a quick recall!

6. Visual Suggestion

6. Visual Suggestion

A flowchart illustrating the conversion of power and time into energy, showing the steps of multiplying power (in watts) by time (in seconds) to reach the total energy consumed (in joules).

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Thermal Properties of Matter).

Question 3 of 5 NDA MCQ

An electric bulb is connected to 220 V generator. The current drawn is 600 mA. What is the power of the bulb?

Detailed Explanation

Correct Answer

✓ Option (a) is correct. The power of the bulb is calculated as 132 W.

📖 Governing Law / Core Concept

The power \( P \) in an electrical circuit is governed by the formula:

\[ P = VI \]

Where:

  • P = Power (in watts, W)
  • V = Voltage (in volts, V)
  • I = Current (in amperes, A)

🧪 Step-by-Step Breakdown

Given voltage \( V = 220 \, \text{V} \) and current \( I = 600 \, \text{mA} = 0.6 \, \text{A} \), we substitute into the power formula:

\[ P = VI = 220 \times 0.6 \]
\[ P = 132 \, \text{W} \]

🔍 Option Analysis

  • Option (b): 13.2 W - Underestimates power based on incorrect current conversion.
  • Option (c): 1320 W - Overestimates power by miscalculating the unit scale.
  • Option (d): 13200 W - Incorrect due to high miscalculation of power.

⚡ Mnemonic / Speed-Run

Remembering the power formula can be simplified as “Power Equals Voltage Times Current”: PE = V x I.

6. Visual Suggestion

A schematic diagram illustrating the electric circuit connected to the bulb, showing voltage (V) and current (I) flowing through the bulb would be beneficial for visual understanding.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Electricity.

Question 4 of 5 NDA MCQ

Which one of the following terms cannot represent electrical power in a circuit?

Detailed Explanation

Correct Answer

✓ Option (b) is correct. \( I^2/R \) does not represent electrical power.

📖 Governing Law / Core Concept

The concept of electric power in a circuit is defined by the following formulas:

  • Power by voltage and current: \( P = VI \)
  • Power in terms of current and resistance: \( P = I^2R \)
  • Power in terms of voltage and resistance: \( P = \frac{V^2}{R} \)

🧪 Step-by-Step Breakdown

The correct expressions for electrical power are based on Ohm's law, where voltage \( V \), current \( I \), and resistance \( R \) relate as:

\[ V = IR \]
1. Using \( P = VI \):
\[ P = (IR)I = I^2R \]
2. Using \( P = \frac{V^2}{R} \):
\[ P = \frac{(IR)^2}{R} = \frac{I^2R}{R} \]
3. Therefore, options A, C, and D are valid representations of electric power. However:
\[ P = I^2/R \] \text{ is not a valid representation.}

🔍 Option Analysis

The options can be analyzed as follows:

  • A: \( VI \) - Correct representation of power.
  • B: \( I^2/R \) - Incorrect. Represents neither direct power computation nor viable transformation.
  • C: \( I^2R \) - Correct representation of power.
  • D: \( \frac{V^2}{R} \) - Correct representation of power.

⚡ Mnemonic / Speed-Run

To remember power formulas:

  • VI = I2R = V2/R - "Voltage, Current, Resistance" for power relationships.

6. Visual Suggestion

6. Visual Suggestion

A circuit diagram showing connected components \( V \), \( I \), and \( R \) with power flow direction. Label all variables clearly to illustrate their relationships.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Electrical Power and Circuits.

Question 5 of 5 NDA MCQ

The cost of energy to operate an industrial refrigerator that consumes 5kW power working 10 hours per day for 30 days will be (Given that the charge per kWh of energy = ₹4) 

Detailed Explanation

Correct Answer

✓ Option (b) is correct. The total cost amounts to ₹6,000.

📖 Governing Law / Core Concept

The cost of energy is derived from the formula: \[ \text{Cost} = \text{Power (kW)} \times \text{Time (h)} \times \text{Cost per kWh} \]

🧪 Step-by-Step Breakdown

Given: - Power \( P = 5 \, \text{kW} \) - Daily Operating Time \( T = 10 \, \text{h} \) - Total Days \( D = 30 \) - Cost per kWh \( C = ₹4 \) Total Energy consumed is: \[ E = P \times T \times D \] Substituting the values: \[ \begin{aligned} E &= 5 \, \text{kW} \times 10 \, \text{h} \times 30 \\ &= 1500 \, \text{kWh} \end{aligned} \] Total cost: \[ \text{Cost} = E \times C = 1500 \, \text{kWh} \times ₹4 \] \[ \text{Cost} = ₹6000 \]

🔍 Option Analysis

- **Option (a) ₹600**: This is underestimated as it considers less operational time or wrong power consumption. - **Option (c) ₹1,200**: Incorrectly calculated energy use or cost. - **Option (d) ₹1,500**: Misinterpretation of total kWh consumed.

⚡ Mnemonic / Speed-Run

To remember the formula, think of **PCT** (Power x Time x Cost).
📊 Visual Diagram Suggestion: A simple flowchart illustrating the relationship between Power, Time, Energy, and Cost. Display arrows indicating the flow from Power and Time to Energy and then to Cost, emphasizing the multiplication involved in calculations.
📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Work, Energy and Power.

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