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Chapter 10 of 11
Important Questions

Electronics and Communication — Important Questions

Tamil Nadu Board · Class 12 · Physics

87 important questions from Electronics and Communication for Tamil Nadu Board Class 12 Physics, with answers. Includes multiple choice questions.

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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87 Questions·
multiple choice

Important Questions from Electronics and Communication

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?

Show answer

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.

+83 more questions on Electronics and Communication (Tamil Nadu Board Class 12 Physics)

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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 Energy Band Diagram and Classification of Materials, Intrinsic and Extrinsic Semiconductors, P-N Junction Diode Formation and Characteristics, Rectification: Half-Wave and Full-Wave Rectifiers. Study these first, then practise questions on each for the Tamil Nadu Board Class 12 board exam.
How many important questions are there in Electronics and Communication?
Super Tutor has 87 practice questions for Electronics and Communication, including multiple choice questions. A sample with answers is on this page.

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