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Friday, 9 October

JEE Main 2023 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 8 April 2023, Shift 2, Physics Q43

Question 43 of 90 in this shift, Physics question 13 of 30, Section A.

For a given transistor amplifier circuit in CE configuration VCC=1V_{CC} = 1 V, RC=1R_C = 1 kΩ\Omega, Rb=100R_b = 100 kΩ\Omega and β=100\beta = 100. Value of base current IbI_b is

The figure, in words

NPN transistor in CE configuration: base connected through R_b = 100 kΩ to battery V_BB (base current I_b marked), collector connected through R_C = 1 kΩ to V_CC = 1 V (collector current I_c marked), emitter current I_e marked
  1. (1)Ib=0.1 μI_b = 0.1\ \muA
  2. (2)Ib=10 μI_b = 10\ \muAOfficial answer
  3. (3)Ib=1.0 μI_b = 1.0\ \muA
  4. (4)Ib=100 μI_b = 100\ \muA

Official answer

Option 2

NTA final key.

Same idea in other shifts

Asked 3× in all
  1. 24 Jun 2022, Shift 1 · Q53A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base…EasyNumerical value
  2. 18 Mar 2021, Shift 1 · Q21An npn transistor operates as a common emitter amplifier with a power gain of 10610^6. The input circuit resistance is 100 Ω\Omega and the…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.