Skip to main content

Assuming forward bias condition (voltage drop 0.7 V per silicon diode), find current through diode D1 in the given circuit. All diodes identical.

Assuming forward bias condition (voltage drop 0.7 V per silicon diode),
find current through diode D1 in the given circuit.

All diodes identicalOptions: A) 20.15 mA B) 11.7 mA C) 17.6 mA D) 18.8 mA

Solution

Apply Kirchhoff's loop rule. Net effective voltage after diode drop: 12 − 0.7 = 11.3 V Using symmetry and branch current division: Total current I ≈ 37.6 mA Current through D1: I₁ = I / 2 I₁ ≈ 18.8 mA Correct Answer: D

Comments

Popular posts from this blog

IAT 2025 Question Paper

IAT 2025 Question Paper Marking Scheme: +4 Correct | -1 Incorrect | 0 Unattempted Q1. (Physics) A ball is projected vertically upward with speed 30 m/s. Neglect air resistance. The time taken to return to the point of projection is: (A) 3 s (B) 4 s (C) 5 s (D) 6 s Show Answer Answer: (D) Total time = 2u/g = 2×30/10 = 6 s. Q2. (Chemistry) Which species has the maximum number of unpaired electrons? (A) Fe²⁺ (B) Fe³⁺ (C) Mn²⁺ (D) Cu²⁺ Show Answer Answer: (C) Mn²⁺ = 3d⁵ configuration → maximum unpaired electrons. Q3. (Mathematics) If sin θ = 3/5 and θ lies in first quadrant, then cos θ is: (A) 4/5 (B) 3/4 (C) 5/4 (D) 2/5 Show Answer Answer: (A) cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25 ⇒ cos θ = 4/5. Q4. (Biology) The functional unit of kidney is: (A) Neuron (B) Alveoli (C) Nephron (D) Glomerulus Show Answer Answer: (C) Nephron is the structural and functional unit of kidney. Q5. (Physics) Escape velocity from Earth is approximately: (A) 7.9 km/s (B) 9.8 km/...
IAT Mock Test 2026 – Part 1 Marking Scheme: +4 Correct | -1 Incorrect | 0 Unattempted Q1. A particle moves in a straight line such that its displacement is given by x = t³ − 6t² + 9t + 4. At what time is its velocity zero? (A) 1 s (B) 2 s (C) 3 s (D) Both A and C Show Answer Correct Answer: (D) v = dx/dt = 3t² − 12t + 9 = 3(t−1)(t−3). Therefore velocity is zero at t = 1 s and 3 s. Q2. The number of stereoisomers possible for a compound with two chiral centers is: (A) 2 (B) 4 (C) 6 (D) 8 Show Answer Correct Answer: (B) Maximum stereoisomers = 2ⁿ, where n = number of chiral centers. Hence 2² = 4. Q3. If z = 1 + i√3, then the principal argument of z is: (A) π/6 (B) π/3 (C) π/2 (D) 2π/3 Show Answer Correct Answer: (B) tan θ = √3/1 = √3, first quadrant ⇒ θ = π/3. Q4. A wire of resistance R is stretched to double its original length. The new resistance becomes: (A) R/2 (B) R (C) 2R (D) 4R Show Answer Correct Answer: (D) On stretching to double length, area...

Two wires A and B have: Length A = 6.0 cm Length B = 5.4 cm Area A = 3.0 × 10⁻⁵ m² Area B = 4.5 × 10⁻⁵ m² Both are stretched by same force producing same extension. If ratio of Young’s modulus is A:B = x : 3, find x

Options: A) 1 B) 4 C) 2 D) 5 Solution Using: Y = (F L) / (A ΔL) Since F and ΔL same: YA / YB = (LA AB) / (LB AA) Substitute: = (6.0 × 4.5) / (5.4 × 3.0) = 27 / 16.2 = 5 / 3 So, x = 5 Correct Answer: D