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Simple electrical circuits

Chapter: Electricity and Magnetism

Question 1 of 5 NDA MCQ

An electric wire of resistance 50 ohm is cut intofive equal wires. These wires are then connectedin parallel. What is the equivalent resistance ofthis combination?

Detailed Explanation

Solution

Option (a) is correct.

Explanation:

\[ R = 50 \, \Omega \]
Each wire's resistance after cutting:
\[ R_{each} = \frac{R}{n} = \frac{50}{5} = 10 \, \Omega \]
For parallel resistances, use:
\[ R_{eq} = \frac{R_{each}}{n} = \frac{10 \, \Omega}{5} = 2 \, \Omega \]
Thus, the equivalent resistance is:
\[ R_{eq} = 2 \, \Omega \]
📌 Hints / Properties Used:
  • Resistance in series: \( R_{total} = R_1 + R_2 + \ldots + R_n \)
  • Resistance in parallel: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \)
📊 Visual Diagram Suggestion:

A circuit diagram showing five parallel resistors, each with a resistance of 10 ohms, converging to a total resistance of 2 ohms. Clearly label each segment and the total circuit layout.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Electricity).

Question 2 of 5 NDA MCQ

Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across wire A, causing power dissipation P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is applied across it causing power dissipation P₁. Which one of the following is the correct relationship between P and P₁?

Detailed Explanation

Solution

Option (b) is correct.

Explanation:

\[ R = \rho \frac{l}{A} \]
For wire A: - Area \( A = \pi r^2 \) - Resistance \( R_A = \rho \frac{l}{\pi r^2} \) For wire B: - Area \( A_B = \pi (2r)^2 = 4\pi r^2 \) - Resistance \( R_B = \rho \frac{2l}{4\pi r^2} = \frac{\rho l}{2\pi r^2} \) The power \( P \) is given by: \[ P = \frac{V^2}{R} \] Power dissipation for wire A: \[ P = \frac{V^2}{R_A} = \frac{V^2}{\rho \frac{l}{\pi r^2}} = \frac{V^2 \cdot \pi r^2}{\rho l} \] Power dissipation for wire B: \[ P_1 = \frac{V^2}{R_B} = \frac{V^2}{\frac{\rho l}{2\pi r^2}} = \frac{2V^2 \cdot \pi r^2}{\rho l} \] Now, we find the relationship between \( P \) and \( P_1 \): \[ \frac{P}{P_1} = \frac{\frac{V^2 \cdot \pi r^2}{\rho l}}{\frac{2V^2 \cdot \pi r^2}{\rho l}} = \frac{1}{2} \implies P = \frac{P_1}{2} \] Hence, the relationship between \( P \) and \( P_1 \) simplifies to \( P = \frac{P_1}{2} \).
📌 Hints / Properties Used:
  • Resistance formula: \( R = \rho \frac{l}{A} \)
  • Power formula: \( P = \frac{V^2}{R} \)
📊 Visual Diagram Suggestion:

Illustrate two wires, with their lengths and radii labeled, and show how resistance and power change between the two wires when the same voltage is applied.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Electricity).

Question 3 of 5 NDA MCQ

Two conducting wires of the same material and of equal length and equal diameter are first connected in parallel and then in series ina circuit across the same potential difference.The ratio of heat produced in parallel and series combination is

Detailed Explanation
✓

1. Correct Answer

Option D is correct. The ratio of heat produced in parallel and series is 1:4.

📖 Core Concept

The heat produced in a conductor is governed by Joule's Law, which relates heat to current, resistance, and time.

💡 3. Explanation

Resistances: Each wire has resistance \( R \). In parallel:

\[ R_P = \frac{R}{2} \]

In series:

\[ R_S = 2R \]

For heat produced:

\[ H \propto P \cdot t \]

Power in series:

\[ P_S = \frac{V^2}{R_S} = \frac{V^2}{2R} \]

Power in parallel:

\[ P_P = \frac{V^2}{R_P} = \frac{2V^2}{R} \]

The ratio of heat produced:

\[ \frac{H_P}{H_S} = \frac{P_P}{P_S} = \frac{2V^2/R}{V^2/(2R)} = \frac{4}{1} \]

Thus, the ratio is 4:1.

❌ 4. Option Analysis

  • • Option A: Incorrect, ratio is 4:1, not 2:1.
  • • Option B: Incorrect, ratio is 4:1, not 4:1.
  • • Option C: Incorrect, ratio is 4:1, not 1:2.
⚡

NDA Speed-Run / Mnemonic Shortcut

Remember: In parallel, the resistance is halved, increasing power and heat generation significantly.

🎯

5. Key Takeaways

Heat in conductors differs significantly under series vs. parallel combinations.

Question 4 of 5 NDA MCQ

If three resistors of 1 Ohm each, connect in parallel to each other, then resultant resistance is

Detailed Explanation
✓

1. Correct Answer

Option B is correct. The resultant resistance is 1/3 Ohm.

📖 Core Concept

In parallel circuits, the total resistance decreases as more resistors are added, calculated by summing the reciprocals of each individual resistor's resistance.

💡 3. Explanation

The formula for equivalent resistance \( R_{eq} \) for resistors in parallel is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
For 3 resistors of 1 Ohm:
\[ R_1 = R_2 = R_3 = 1\, Ohm \]
Thus,
\[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \]
Therefore,
\[ R_{eq} = \frac{1}{3}\, Ohm \]

💡 Hints & Visual Guide

📌 Hints / Properties Used:

  • Use \( R_{eq} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \) for parallel resistances.
  • Understand that the equivalent resistance reduces when resistors are added in parallel.

📊 Visual Diagram Suggestion:

[Illustration showing three 1 Ohm resistors connected in parallel, indicating the effective resistance computation as \( \frac{1}{3}\, Ohm \)]

Question 5 of 5 NDA MCQ

An incandescent electric bulb converts 20%of its power consumption into light and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 W and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity chargeis ₹ 5/unit, then which among the followingis the correct amount of the money spent on producing light?

Detailed Explanation

Solution

Option (d) is correct.

Explanation:

\[ P = I^2R \]
Here, \( I = 2 \, \text{A} \) and \( R = 200 \, \Omega \).
\[ P = (2)^2 \times 200 = 800 \, \text{W} \]
Total power consumption in kW:
\[ P = 0.8 \, \text{kW} \]
Only 20% of this power is used for light:
\[ P_{\text{light}} = 0.8 \times \frac{20}{100} = 0.16 \, \text{kW} \]
Energy consumed over 10 hours:
\[ E = P_{\text{light}} \times t = 0.16 \, \text{kW} \times 10 \, \text{h} = 1.6 \, \text{kWh} = 1.6 \, \text{units} \]
Cost of electricity at ₹5/unit:
\[ \text{Cost} = 1.6 \times 5 = ₹8 \]
📌 Hints / Properties Used:
  • Power formula: \( P = I^2R \)
  • Energy calculation: \( E = P \times t \)
  • Cost: \( \text{Cost} = \text{Units} \times \text{Rate} \)
📊 Visual Diagram Suggestion:

A schematic circuit diagram representing the incandescent bulb, indicating the current and resistance values, along with power dissipation percentages would enhance comprehension.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 4 (Electromagnetic Induction).

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