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Simple electrical circuits Questions

Chapter: Electricity and Magnetism | Total Available: 19 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

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?

Step-by-Step Logic & Solution

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 10 Authentic UPSC NDA Question

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?

Step-by-Step Logic & Solution

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 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution
✓

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 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution
✓

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 10 Authentic UPSC NDA Question

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?

Step-by-Step Logic & Solution

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).

Question 6 of 10 Authentic UPSC NDA Question

For an electric circuit given below, the correctcombination of voltage (V) and current (I) is

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

\[ V = I_1 R_1 \]
Given \( I_1 = 3A \):
\[ V = 3 \times 100 = 300V \]
Current through \( R_2 \):
\[ I_2 = \frac{V}{R_2} = \frac{300}{200} = 1.5 \, A \]
Current through \( R_3 \):
\[ I_3 = \frac{V}{R_3} = \frac{300}{300} = 1 \, A \]
Net current \( I \):
\[ I = I_1 + I_2 + I_3 = 3 + 1.5 + 1 = 5.5 \, A \]
📌 Hints / Properties Used:
  • Ohm's Law: \( V = IR \)
  • Current in parallel: \( I_{total} = I_1 + I_2 + I_3 \)
📊 Visual Diagram Suggestion:

A schematic showing the parallel connection of resistors with wires indicating current flows and labeled voltage sources to clarify voltage drops and current distribution.

📖 Factual Verification & Reference:

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

Question 7 of 10 Authentic UPSC NDA Question

In which one of the following situations, the bulb would glow the most? (Consider allbatteries are the same.)

Step-by-Step Logic & Solution

Correct Answer

✓ Option (d) is correct. The two bulbs connected in parallel with two cells in series will glow brighter.

📖 Governing Law / Core Concept

The brightness of a bulb is directly proportional to the power it receives, represented by the formula:
\[ P = \frac{V^2}{R} \]
Where \( P \) is the power (in Watts), \( V \) is the voltage (in Volts), and \( R \) is the resistance (in Ohms).

🧪 Step-by-Step Breakdown

For bulbs connected in parallel, the voltage across each bulb remains constant. In this case, when two identical batteries are connected in series, the total voltage is:
\[ V_{\text{total}} = V_1 + V_2 \]
Given identical batteries, the total voltage becomes:
\[ V_{\text{total}} = 2V \]
In parallel, the power in each bulb is:
\[ P = \frac{(2V)^2}{R} = \frac{4V^2}{R} \]
Thus, each bulb will glow with maximum brightness, significantly brighter compared to series configurations.

🔍 Option Analysis

Options (a), (b), and (c) are series configurations, limiting voltage and power, resulting in dimmer bulbs.

⚡ Mnemonic / Speed-Run

For maximum brightness in circuits: Batteries in Series + Bulbs in Parallel = Highest Glow.

6. Visual Suggestion

Illustrate a circuit diagram showing two bulbs connected in parallel with two batteries in series, clearly indicating the equal voltage across each bulb for enhanced comprehension.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 9 (Electrical Circuits).

Question 8 of 10 Authentic UPSC NDA Question

Two resistances of 5.0 Ωand 7.0Ω are connectedin series and the combination is connectedin parallel with a resistance of 36.0Ω. Theequivalent resistance of the combination ofthree resistors is:

Step-by-Step Logic & Solution

Solution

Option (c) is correct.

Explanation:

\[ R_s = R_1 + R_2 \]
For \( R_1 = 5\, \Omega \) and \( R_2 = 7\, \Omega \):
\[ R_s = 5 + 7 = 12\, \Omega \]
Next, for parallel connection with \( R_3 = 36\, \Omega \):
\[ \frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{R_3} \]
Substituting values:
\[ \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{36} \]
Finding a common denominator (36):
\[ \frac{1}{R_{eq}} = \frac{3}{36} + \frac{1}{36} = \frac{4}{36} = \frac{1}{9} \]
Taking the reciprocal:
\[ R_{eq} = 9\, \Omega \]
📌 Hints / Properties Used:
  • Series resistance: \( R_s = R_1 + R_2 \)
  • Parallel resistance: \( \frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{R_3} \)
📊 Visual Diagram Suggestion:

Illustrate a series connection of two resistors (5.0 Ω and 7.0 Ω) followed by a parallel connection to a third resistor (36.0 Ω). Clearly label the resistances to aid visual comprehension.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Current Electricity.

Question 9 of 10 Authentic UPSC NDA Question

Consider the following electric circuit:

Step-by-Step Logic & Solution

Correct Answer

✓ Option (A) is correct. The total current in the circuit is 1 A.

📖 Governing Law / Core Concept

The circuit analysis is based on Ohm's Law, expressed as:

\[ I = \frac{V}{R_{eq}} \]

Where \( I \) is the current (A), \( V \) is the voltage (V), and \( R_{eq} \) is the equivalent resistance (Ω).

🧪 Step-by-Step Breakdown

Given the circuit configuration, analyze the resistances:

\[ R_{eq} = R_1 + R_{parallel} \]

Calculating the parallel resistance of \( 10Ω \) and \( 0Ω \):

\[ \begin{aligned} \frac{1}{R_{parallel}} &= \frac{1}{10} + \frac{1}{0} \\ R_{parallel} &= 0Ω \end{aligned} \]
\[ R_{eq} = 5 + 0 + 5 = 10Ω \]

Now applying Ohm's Law:

\[ I = \frac{10V}{10Ω} = 1 A \]

📌 Hints / Properties Used:

  • Ohm's Law: \( V = I \times R \)
  • Equivalent Resistance in series and parallel combinations.

📊 Visual Diagram Suggestion:

A circuit diagram illustrating the total voltage source (10 V), the series and parallel configurations of resistors (5Ω, 10Ω, and 0Ω), and the current flow direction, highlighting the computation steps to find \( R_{eq} \) and \( I \).

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 6 (Current Electricity).

Question 10 of 10 Authentic UPSC NDA Question

An infinite combination of resistors, eachhaving resistance R = 4 W, is given below.What is the net resistance between the points Aand B? (Each resistance is of equal value, R = 4.

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

\[ R = 4 \, \Omega \]
\[ R_{net} = R + \frac{R}{n} \; \text{(for infinite resistors)} \]
\[ R_{net} = \frac{R(1 + \frac{1}{n})}{1 - \frac{1}{n}} \; \text{(geometric series)} \]
\[ = \frac{4(1 + 1)}{1 - 1} = \infty \; \text{(not converging)} \]
\[ R_{AB} = \frac{2 \times 4}{2 + 4} = 2 + 2\sqrt{5} \, \Omega \]
\[ R_{total} = 2 + 2\sqrt{5} \, \Omega \]
📌 Hints / Properties Used:
  • Series sum: \( R_{series} = R_1 + R_2 + ... \)
  • Parallel sum: \( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + ... \)
  • Geometric series: \( S = a \frac{1 - r^{n}}{1 - r} \)
📊 Visual Diagram Suggestion:

Illustrate the infinite series connection of resistors A to B, labeling each resistor \( R = 4 \, \Omega \) clearly.

📖 Factual Verification & Reference:

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

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