For a given transistor amplifier circuit in CE configuration $V_{CC} = 1$ V, $R_C = 1$ k$\Omega$, $R_b = 100$ k$\Omega$ and $\beta = 100$. Value of base current $I_b$ is

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

V_{CC} = 1

R_C = 1

\Omega

R_b = 100

\Omega

and

\beta = 100

. Value of base current

I_b

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) $I_b = 0.1\ \mu$ A
(2) $I_b = 10\ \mu$ AOfficial answer
(3) $I_b = 1.0\ \mu$ A
(4) $I_b = 100\ \mu$ A

Official answer

Option 2

NTA final key.

Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits
Topic: Transistors
Idea tested: CE Amplifier Voltage Gain

Same idea in other shifts

Asked 7× in all

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 108 of the 174 JEE Main shifts held from 2017 to 2026.