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.
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Important Questions from Electronics and Communication
For a transistor in common base configuration, α = 0.98 and IE = 2 mA. What is the base current IB?
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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.
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?
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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.
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.
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)
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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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