Junction Diode Numerical Class-12 Nootan ISC Physics Solution Ch-32. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Junction Diode Numerical Class-12 Nootan ISC Physics Solution Ch-32
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-32 | Junction Diode |
| Topics | Numericals on Junction Diode |
| Academic Session | 2025-2026 |
Numericals on Junction Diode
Class-12 Nootan ISC Physics Solution Ch-32
Que-1: A p-n junction diode can withstand currents upto a maximum of 10 mA. A resistor R = 200 Ω is connected in series with it. The diode, when forward-biased, has a potential drop of 0.5 V across it which is assumed to be independent of current. Find the maximum voltage of the battery used to forward bias the diode.
Ans- i = 10 mA = 10 x 10^-3 A
R = 200 Ω , V = 0.5 V
Using Ohm’s law
i R + 0.5 V = Vmax
10 x 10^-3 x 200 + 0.5 = Vmax
2000 x 10^-3 + 0.5 = Vmax
2.0 + 0.5 = Vmax
Vmax = 2.5 V
Que-2: When the voltage drop across a p-n junction diode is increased from 0.65 V to 0.70 V, the change in the diode current is 5 mA. What is the dynamic resistance of the diode?
Ans- Dynamic resistance
R = ΔV/Δi
= 0.70 – 0.65 / 5 x 10^-3
= 0.05 x 1000 / 5
= 50/5
= 10 Ω
Que-3: A p-n junction diode is connected to a series resistor of 100 Ω and a 4.5 V DC battery. The barrier potential at the junction is 0.5 V. Find the current in the circuit.
Ans- R = 100 Ω , V = 4.5 Volt , barrier potential = 0.5 V
i R + 0.5 = 4.5
i x 100 = 4.5 – 0.5
i x 100 = 4.0
i = 4/100 = 40/1000
i = 40 x 10^-3
i = 40 mA
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