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Chapter 10 of 11
Practice Quiz

Electronics and Communication

Tamil Nadu Board · Class 12 · Physics

Practice quiz for Electronics and Communication — Tamil Nadu Board Class 12 Physics. MCQs and questions with answers to test your preparation.

87 questions30 flashcards5 concepts

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A graph showing the current (I) versus voltage (V) characteristics of a p-n junction diode. Include both forward bias (exponential increase after knee voltage) and reverse bias (small reverse saturati
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Quick Quiz: Electronics and Communication

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1

In an NPN transistor connected in common emitter configuration, the base current is 50 µA and the current gain β = 100. What is the collector current?

2

The forbidden energy gap for silicon at room temperature is 1.1 eV. What does this value tell us about silicon?

3

A silicon diode is forward biased with a supply voltage of 10 V and a series resistance of 1 kΩ. Taking the barrier potential of silicon as 0.7 V, what is the current through the circuit?

4

In a full wave rectifier, what is the efficiency compared to a half wave rectifier?

87 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

For a transistor in common base configuration, α = 0.98 and IE = 2 mA. What is the base current IB?

Show answer

0.04 mA

Step 1: Find IC using the formula α = IC/IE. So IC = α × IE = 0.98 × 2 = 1.96 mA. Step 2: Apply Kirchhoff's current law: IE = IB + IC. Step 3: Rearrange: IB = IE − IC = 2 − 1.96 = 0.04 mA. Option B (1.96 mA) is the collector current, not base current. Option C (2.04 mA) is wrongly adding instead of subtracting. Option D uses incorrect formula.

2multiple choice
1 marks

A Zener diode has a breakdown voltage of 6 V. It is connected in a circuit with a series resistance of 500 Ω and an input voltage of 12 V. What is the current through the series resistor?

Show answer

12 mA

Step 1: In the Zener voltage regulator circuit, the Zener diode maintains a constant voltage equal to its breakdown voltage VZ = 6 V across itself (and the load). Step 2: Voltage across the series resistor = Input voltage − VZ = 12 − 6 = 6 V. Step 3: Current through series resistor I = V/R = 6/500 = 0.012 A = 12 mA. Option B (24 mA) uses full 12 V instead of the voltage drop across Rs. Option C (6 mA) uses 3 V as drop. Option D divides incorrectly.

3multiple choice
1 marks

Which of the following logic gate combinations is known as a 'Universal Gate' and can be used to construct ANY other logic gate?

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NAND gate alone or NOR gate alone

Step 1: A universal gate is one from which all other basic gates (AND, OR, NOT) can be constructed. Step 2: NAND gate: NOT = A NAND A, AND = (A NAND B) NAND (A NAND B), OR = (A NAND A) NAND (B NAND B). Similarly NOR can be used to make all basic gates. Step 3: AND + OR combination is insufficient because you cannot make a NOT gate from them alone. Ex-OR is a special purpose gate. NOT gate alone can only invert, cannot create AND or OR logic.

4multiple choice
1 marks

An LED is made of GaAsP semiconductor with a forbidden energy gap of 1.875 eV. What is the wavelength of light emitted? (h = 6.6 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

Show answer

660 nm (Red light)

Step 1: Use the photon energy formula: Eg = hc/λ, so λ = hc/Eg. Step 2: Convert Eg to Joules: Eg = 1.875 × 1.6 × 10⁻¹⁹ = 3.0 × 10⁻¹⁹ J. Step 3: Calculate λ = (6.6 × 10⁻³⁴ × 3 × 10⁸) / (3.0 × 10⁻¹⁹) = 19.8 × 10⁻²⁶ / 3.0 × 10⁻¹⁹ = 6.6 × 10⁻⁷ m = 660 nm. Step 4: 660 nm falls in the red region of the visible spectrum. Option B (330 nm) results from forgetting to convert eV to J. Option C doubles the wavelength. Option D is the green region (~550 nm) corresponding to a different energy gap.

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Frequently Asked Questions

What are the important topics in Electronics and Communication for Tamil Nadu Board Class 12 Physics?
Key topics in Electronics and Communication include Mind map showing the hierarchical structure of electronics and communication topics covered in this chapter, Flowchart showing how isolated atoms' orbitals combine to form energy bands in solids, with different band structures for insulators, semiconductors, and conductors, State diagram showing material classification based on band structure and resulting electrical properties. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Electronics and Communication — Tamil Nadu Board Class 12 Physics?
Understand the core concepts first, then work through the 87 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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